AP: First Term 3, Sum of First 4 Terms = 1/5 of Next 4 — Sum of 20 Terms
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.
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Limit of Σ(k³+6k²+11k+5)/(k+3)! From k=1 to n as n→∞
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.
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AP: a1+(a5+a10+…+a2020)+a2024=2233, Find Sum to 2024 Terms
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.
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Sum of 1 + 6 + 9(1²+2²+3²)/7 + 12(1²+2²+3²+4²)/9 + … Up to 15 Terms
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.
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G.P. of Increasing Positive Terms, a1a5=28 and a2+a4=29, Find a6
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.
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Sum to 10 Terms of the Series r/(1+r²+r⁴), JEE Main 2023
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.
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How to Find the Sum of 7 + 77 + 777 + … Up to n Terms
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.
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If Sn=1/2+1/6+1/12+… to n Terms, A.P. Sum is √(2026·S2025)
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.