Two APs: s1=t1≠0, s2=2t2 — Find (s2−s1)/(t2−t1)
Two arithmetic sequences share a first term and satisfy s2=2t2. Given the sum of 10 terms of one equals the sum of 15 terms of the other, find (s2−s1)/(t2−t1).
Two arithmetic sequences share a first term and satisfy s2=2t2. Given the sum of 10 terms of one equals the sum of 15 terms of the other, find (s2−s1)/(t2−t1).
Prove that if the sum of the first m terms of an AP equals the sum of the first n terms (m≠n), then the sum of the first m+n terms must be zero.
An AP has first term 3, and the sum of its first 4 terms equals 1/5 of the sum of the next 4 terms. Find the sum of the first 20 terms.
Evaluate the infinite series limit of (k³+6k²+11k+5)/(k+3)! by rewriting the numerator in terms of j=k+3 to reveal a telescoping factorial pattern.
For an AP a1,…,a2024 with a1+(a5+a10+…+a2020)+a2024=2233, find a1+a2+…+a2024 by summing the selected terms as their own smaller AP.
The sum of the series 1+6+9(1²+2²+3²)/7+12(1²+2²+3²+4²)/9+15(1²+…+5²)/11+… up to 15 terms is: (1)7820 (2)7520 (3)7830 (4)7510.
Let a1,a2,a3,… be a G.P. of increasing positive terms. If a1a5=28 and a2+a4=29, then a6 is equal to: (A) 628 (B) 812 (C) 526 (D) 784.
The sum to 10 terms of the series 1/(1+1²+1⁴) + 2/(1+2²+2⁴) + 3/(1+3²+3⁴) + … is (A) 56/111 (B) 58/111 (C) 55/111 (D) 50/111.
A step-by-step derivation of the sum of the series 7 + 77 + 777 + … up to n terms, using the repunit-pattern trick to write each term as 7(10^n-1)/9.
A JEE Main 2025 problem combining a telescoping series with an arithmetic progression, solved by simplifying Sn first, then using it to pin down the AP.