Mathematics · 2026
JEE Main · 2 April 2026, Shift 2 · Q4
Let a_1, a_2, a_3, …. be an A.P. and g_1= a_1, g_2, g_3, …. be an increasing G.P. If a_1= a_2+ g_2=1 and a_3+ g_3=4, then a_10+ g_5 is equal to:
Let $\displaystyle \mathrm{a}_1, \mathrm{a}_2, \mathrm{a}_3, \ldots$. be an A.P. and $\displaystyle \mathrm{g}_1=\mathrm{a}_1, \mathrm{~g}_2, \mathrm{~g}_3, \ldots$. be an increasing G.P. If $\displaystyle \mathrm{a}_1=\mathrm{a}_2+\mathrm{g}_2=1$ and $\displaystyle \mathrm{a}_3+\mathrm{g}_3=4$, then $\displaystyle \mathrm{a}_{10}+\mathrm{g}_5$ is equal to :
Official answer
From NTA’s final answer key for this paper.
(4)
$\displaystyle 55$
More from Progressions
- Let A be the set of first 101 terms of an A.P., whose first term is 1 and the common difference is 5 and let…2026
- The first term of an A.P. of 30 non-negative terms is 10/3. If the sum of this A.P. is the cube of its last…2026
- If Σ_k=1^n a_k=6 n^3, then Σ_k=1^6((a_k+1-a_k)/36)^2 is equal to ____.2026
- Σ_n=1^10(528/(n(n+1)(n+2))) is equal to:2026
- The sum of the first ten terms of an A.P. is 160 and the sum of the first two terms of a G.P. is 8. If the…2026
- Let the sum of the first n terms of an A.P. be 3 n^2+5 n. Then the sum of squares of the first 10 terms of…2026
- The value of 1^3-2^3+3^3-…+15^3 is:2026
- The sum 1+1/2(1^2+2^2)+1/3(1^2+2^2+3^2)+…. upto 10 terms is equal to:2026
JEE Main 2026 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.