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AP and Increasing GP: Find a10 + g5 = 55 | JEE Main 2026

JEE Maths question with a full step-by-step solution.

Question
Let a1,a2,a3,…a_1,a_2,a_3,\ldots be an A.P. and g1=a1,g2,g3,…g_1=a_1,g_2,g_3,\ldots be an increasing G.P. If a1=a2+g2=1a_1=a_2+g_2=1 and a3+g3=4a_3+g_3=4, then a10+g5a_{10}+g_5 is equal to
A8181
B7676
C6262
D5555correct
Solution
Step 1: Since a1=1a_1=1 and g1=a1=1g_1=a_1=1, write
A.P.: 1, a2, a3,… (common difference d),G.P.: 1, g2, g3,… (common ratio r).\text{A.P.}:\ 1,\ a_2,\ a_3,\ldots\ (\text{common difference } d),\qquad \text{G.P.}:\ 1,\ g_2,\ g_3,\ldots\ (\text{common ratio } r).
 Step 2: The condition a2+g2=1a_2+g_2=1:
(1+d)+r=1 ⇒ d+r=0.(1+d)+r=1\ \Rightarrow\ d+r=0.
 Step 3: The condition a3+g3=4a_3+g_3=4:
(1+2d)+r2=4 ⇒ 2d+r2=3.(1+2d)+r^2=4\ \Rightarrow\ 2d+r^2=3.
 Step 4: Substitute d=−rd=-r:
−2r+r2=3 ⇒ r2−2r−3=0 ⇒ r=3 or r=−1.-2r+r^2=3\ \Rightarrow\ r^2-2r-3=0\ \Rightarrow\ r=3\ \text{or}\ r=-1.
 Reject r=−1r=-1 (the G.P. is increasing), so r=3r=3 and d=−3d=-3. Step 5:
a10=1+9d=1+9(−3)=−26,g5=1⋅r4=34=81.a_{10}=1+9d=1+9(-3)=-26,\qquad g_5=1\cdot r^4=3^4=81.
 Step 6: $a10+g5=−26+81=55.a_{10}+g_5=-26+81=55. Correct answer: (4)
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