Skip to content
dronacharya.ai
← JEE Main 2022 · 30 Jul · Shift 2
MathematicsSingle correct+4 / -1Question 67 of 90

JEE Main 2022 · 30 Jul · Shift 2, Mathematics question 67

For |α| ≥ 1, let 54−2α5^{4-2\alpha}, 63, 52α−15^{2\alpha-1} be in A.P. with the common difference d. Let Sn_n denote the sum of the first n terms of an A.P., whose first term is α2^2 and the common difference is d. If S30_{30} − S15_{15} = 30k, then k is equal to

Choose one option
Answer and worked solution

Official answer: C

Find α. 54−2α+52α−1=1265^{4-2\alpha}+5^{2\alpha-1}=126; with t=52αt=5^{2\alpha}, t2−630t+3125=0t^2-630t+3125=0, so t=625t=625 (α=2\alpha=2) or t=5t=5 (α=12\alpha=\frac12, rejected). The A.P. 1, 63, 125 has d=62d=62.

Sum of terms 16 to 30 with first term 4: 15×4+62(15+16+⋯+29)=60+62×330=20520=30k15\times4+62(15+16+\dots+29)=60+62\times330=20520=30k, so k=684k=684.

(C)