NCERT Exemplar Solutions Class 10 Maths Arithmetic Progressions — Exercise 5.4

NCERT Exemplar Problems, Class 10 Mathematics — Arithmetic Progressions, Exercise 5.4. All 10 questions solved.

The long-answer exercise is where the AP formulas meet situations that have to be modelled first. Three patterns account for all ten questions.

Counting the multiples in a range. Questions 2 and 5 both ask for the sum of the multiples of some number inside an interval. The method is always: find the first and last such multiple, count them with $l = a + (n-1)d$, then sum with $\frac{n}{2}(a+l)$. For “multiples of 2 or 5″, add the two sums and subtract the multiples of $10$, which have been counted twice.

Ratios of terms. Question 6 gives a ratio and asks for others. The trick is that a ratio condition determines $a$ in terms of $d$ — here $a = 4d$ — and everything else then cancels.

Both ends of a range. Question 10’s flags and Question 9’s instalments both need the situation modelled before any formula applies, and in both the modelling is the hard part.

Key insight. “Between 100 and 200” and “from 1 to 500” are different ranges, and the exemplar sets both in the same question (2(i) against 2(ii)) to see whether you noticed. Write down the first and last member of the list explicitly before counting anything.

Question 1

The sum of the first five terms of an AP and the sum of the first seven terms of the same AP is $167$. If the sum of the first ten terms of this AP is $235$, find the sum of its first twenty terms.

Solution. Write each sum in $a$ and $d$:

$$S_5 = \frac{5}{2}\left(2a + 4d\right) = 5a + 10d, \qquad S_7 = \frac{7}{2}\left(2a + 6d\right) = 7a + 21d$$

$$S_5 + S_7 = 12a + 31d = 167$$

$$S_{10} = \frac{10}{2}\left(2a + 9d\right) = 10a + 45d = 235 \implies 2a + 9d = 47$$

Multiply the second by $6$ to match the $a$ terms:

$$12a + 54d = 282$$

and subtract the first: $23d = 115$, so $d = 5$, and then $2a = 47 – 45 = 2$, giving $a = 1$.

$$S_{20} = \frac{20}{2}\left[2(1) + 19(5)\right] = 10(97) = 970$$

Question 2

Find the

(i) sum of those integers between $1$ and $500$ which are multiples of $2$ as well as of $5$

Multiples of both $2$ and $5$ are multiples of $10$. Between $1$ and $500$ means strictly inside, so the list runs $10$, $20$, …, $490$.

$$490 = 10 + (n-1)10 \implies n = 49$$

$$S = \frac{49}{2}(10 + 490) = \frac{49}{2}(500) = 12250$$

(ii) sum of those integers from $1$ to $500$ which are multiples of $2$ as well as of $5$

“From $1$ to $500$” includes $500$ itself, so the list is $10$, $20$, …, $500$ with $n = 50$:

$$S = \frac{50}{2}(10 + 500) = 25(510) = 12750$$

The two answers differ by exactly $500$ — which is the whole point of setting the same question twice.

(iii) sum of those integers from $1$ to $500$ which are multiples of $2$ or $5$

Adding the two sums would count the multiples of $10$ twice, so subtract them once:

$$\text{multiples of }2: \quad 2 + 4 + \dots + 500 = \frac{250}{2}(2 + 500) = 62750$$ $$\text{multiples of }5: \quad 5 + 10 + \dots + 500 = \frac{100}{2}(5 + 500) = 25250$$ $$\text{multiples of }10: \quad 12750 \ \text{(from part (ii))}$$

$$S = 62750 + 25250 – 12750 = 75250$$

Question 3

The eighth term of an AP is half its second term and the eleventh term exceeds one third of its fourth term by $1$. Find the $15$th term.

Solution. Translate each sentence into an equation.

$$a_8 = \frac{a_2}{2} \implies a + 7d = \frac{a + d}{2} \implies 2a + 14d = a + d \implies a = -13d$$

$$a_{11} = \frac{a_4}{3} + 1 \implies a + 10d = \frac{a + 3d}{3} + 1 \implies 3a + 30d = a + 3d + 3 \implies 2a + 27d = 3$$

Substituting $a = -13d$ into the second:

$$-26d + 27d = 3 \implies d = 3, \qquad a = -39$$

$$a_{15} = -39 + 14(3) = -39 + 42 = 3$$

Question 4

An AP consists of $37$ terms. The sum of the three middle most terms is $225$ and the sum of the last three is $429$. Find the AP.

Solution. With $37$ terms the middle one is the $19$th, so the three middle-most are the $18$th, $19$th and $20$th. Their sum is three times the middle one:

$$3a_{19} = 225 \implies a_{19} = 75$$

The last three are the $35$th, $36$th and $37$th, whose middle is the $36$th:

$$3a_{36} = 429 \implies a_{36} = 143$$

Now the difference of two terms:

$$a_{36} – a_{19} = 17d = 143 – 75 = 68 \implies d = 4$$

and from $a_{19} = a + 18d = 75$: $\ a = 75 – 72 = 3$.

The AP is $3$, $7$, $11$, $15$, …

Question 5

Find the sum of the integers between $100$ and $200$ that are

(i) divisible by $9$

The first multiple of $9$ above $100$ is $108$ ($9 \times 12$); the last below $200$ is $198$ ($9 \times 22$). So the list is $108$, $117$, …, $198$:

$$198 = 108 + (n-1)9 \implies 9(n-1) = 90 \implies n = 11$$

$$S = \frac{11}{2}(108 + 198) = \frac{11}{2}(306) = 1683$$

(ii) not divisible by $9$

Take everything and remove what part (i) counted. The integers strictly between $100$ and $200$ are $101$ to $199$, which is $99$ numbers:

$$101 + 102 + \dots + 199 = \frac{99}{2}(101 + 199) = \frac{99}{2}(300) = 14850$$

$$\text{not divisible by } 9 = 14850 – 1683 = 13167$$

Subtracting is far quicker than trying to sum the non-multiples directly — they do not form an AP.

Question 6

The ratio of the $11$th term to the $18$th term of an AP is $2 : 3$. Find the ratio of the $5$th term to the $21$st term, and also the ratio of the sum of the first five terms to the sum of the first $21$ terms.

Solution. Turn the given ratio into a relation between $a$ and $d$:

$$\frac{a + 10d}{a + 17d} = \frac23 \implies 3(a + 10d) = 2(a + 17d) \implies 3a + 30d = 2a + 34d \implies a = 4d$$

Now every other quantity can be written in $d$ alone.

Ratio of the 5th to the 21st term:

$$\frac{a + 4d}{a + 20d} = \frac{4d + 4d}{4d + 20d} = \frac{8d}{24d} = \frac13$$

Ratio of the sums:

$$\frac{S_5}{S_{21}} = \frac{\frac52\left(2a + 4d\right)}{\frac{21}{2}\left(2a + 20d\right)} = \frac{5\left(8d + 4d\right)}{21\left(8d + 20d\right)} = \frac{5 \times 12d}{21 \times 28d} = \frac{60}{588} = \frac{5}{49}$$

Question 7

Show that the sum of an AP whose first term is $a$, the second term $b$ and the last term $c$, is equal to

$$\frac{(a+c)(b+c-2a)}{2(b-a)}$$

Solution. The second term is $b$, so the common difference is

$$d = b – a$$

The last term is $c$, so the number of terms comes from $c = a + (n-1)d$:

$$n – 1 = \frac{c – a}{b – a} \implies n = \frac{c-a}{b-a} + 1 = \frac{(c-a) + (b-a)}{b-a} = \frac{b + c – 2a}{b – a}$$

Now use the sum formula in its first-and-last-term form, since both ends are known:

$$S = \frac{n}{2}\left(a + c\right) = \frac{1}{2} \cdot \frac{b+c-2a}{b-a} \cdot (a + c) = \frac{(a+c)(b+c-2a)}{2(b-a)}$$

which is the required expression. (The condition $b \ne a$ is implicit — an AP with $d = 0$ has no well-defined number of terms from its last value.)

Question 8

Solve the equation $\ -4 + (-1) + 2 + \dots + x = 437$.

Solution. The left side is an AP with $a = -4$ and $d = 3$, ending at $x$. Write the sum in terms of the number of terms $n$:

$$S_n = \frac{n}{2}\left[-8 + 3(n-1)\right] = \frac{n(3n – 11)}{2} = 437$$

$$3n^2 – 11n – 874 = 0$$

$D = 121 + 4(3)(874) = 121 + 10488 = 10609 = 103^2$, so

$$n = \frac{11 \pm 103}{6} = 19 \ \text{ or } \ -\frac{46}{3}$$

A count of terms must be a positive integer, so $n = 19$, and

$$x = a_{19} = -4 + 18(3) = -4 + 54 = 50$$

Question 9

Jaspal Singh repays his total loan of ₹118000 by paying every month starting with the first instalment of ₹1000. If he increases the instalment by ₹100 every month, what amount will be paid by him in the $30$th instalment? What amount of loan does he still have to pay after the $30$th instalment?

Solution. The instalments form an AP with $a = 1000$ and $d = 100$.

The 30th instalment:

$$a_{30} = 1000 + 29(100) = 1000 + 2900 = ₹3900$$

Amount paid in thirty instalments:

$$S_{30} = \frac{30}{2}\left[2(1000) + 29(100)\right] = 15\left[2000 + 2900\right] = 15 \times 4900 = ₹73500$$

Loan remaining:

$$118000 – 73500 = ₹44500$$

Question 10

The students of a school decided to beautify the school on the Annual Day by fixing colourful flags on the straight passage of the school. They have $27$ flags to be fixed at intervals of every $2$ m. The flags are stored at the position of the middle most flag. Ruchi was given the responsibility of placing the flags. Ruchi kept her books where the flags were stored. She could carry only one flag at a time. How much distance did she cover in completing this job and returning back to collect her books? What is the maximum distance she travelled carrying a flag?

Solution. With $27$ flags the middle one is the 14th, and that is where Ruchi stands. So there are $13$ flags to her left and $13$ to her right, at distances

$$2, \ 4, \ 6, \ \dots, \ 26 \text{ metres}$$

on each side. The 14th flag is planted where she stands, costing no travel.

For each flag she walks out and back, so a flag at distance $s$ costs $2s$. On one side the total is

$$2\left(2 + 4 + \dots + 26\right) = 2 \times \frac{13}{2}(2 + 26) = 2 \times 182 = 364 \text{ m}$$

using $S_{13} = \frac{13}{2}(2 + 26) = 182$ for the distances themselves. The other side is identical, so

$$\text{total} = 364 + 364 = 728 \text{ m}$$

She finishes back at her books, so no extra return trip is needed.

The maximum distance carrying a flag is the one-way trip to the furthest flag, the 13th on either side:

$$26 \text{ m}$$

Common mistakes

• Treating “between 1 and 500” as “from 1 to 500”. Question 2 parts (i) and (ii) differ only in this, and the answers differ by $500$. Write out the first and last member before counting.
• Adding the two sums in Question 2(iii) without subtracting. Multiples of $10$ are multiples of both $2$ and $5$, so they appear in each list; inclusion–exclusion requires removing one copy.
• Summing the non-multiples directly in Question 5(ii). They are not an AP. Subtracting from the total is the only sensible route, and the hint in the book says so.
• Taking the middle-most three as terms 19, 20, 21 in Question 4. With $37$ terms the middle is the $19$th, so the three around it are $18$, $19$, $20$.
• Forgetting the return journey in Question 10. Each flag costs twice its distance, because Ruchi carries one at a time and walks back for the next. Forgetting it halves the answer to $364$.
• Counting the middle flag as a trip in Question 10. It is planted where she stands, at zero distance — which is also why $27$ flags give only $13$ on each side.
• Solving Question 6 for $a$ and $d$ separately. A single ratio cannot determine both; it determines $a = 4d$, and every requested ratio then has $d$ cancel. Trying to pin down numerical values leads nowhere.
• Rounding in Question 8. $10609$ is exactly $103^2$, so the roots are exact. A decimal square root here suggests an arithmetic slip earlier.

Practise next

• Exemplar Exercise 5.3 Q19–35 — shorter versions of these sums, including the double-answer question.
• Exemplar Exercise 5.3 Q1–18 for the term work Questions 3, 4 and 6 above assume.
• NCERT textbook Exercise 5.3 — the book’s own sum problems, including instalments and prize money of the same shape as Question 9.
• NCERT textbook Exercise 5.4 for the optional harder set, where the flag-and-distance modelling of Question 10 also appears.