JEE Main 2024 (Online) 31st January Evening ShiftQuestion 1
Let $2^{\text{nd}}$, $8^{\text{th}}$ and $44^{\text{th}}$ terms of a non-constant A.P. be respectively the $1^{\text{st}}$, $2^{\text{nd}}$ and $3^{\text{rd}}$ terms of a G.P. If the first term of the A.P. is $1$, then the sum of its first $20$ terms is equal to -
Sequence and Series · Mixed Progressions (AP-GP)
Solution
Formulas used. (1) $a_n=a+(n-1)d$ and $S_n=\frac n2\left[2a+(n-1)d\right]$(2) $x,y,z$ in G.P. $\Rightarrow y^2=xz$Step 1 - The three terms. With $a=1$: $$a_2=1+d,\qquad a_8=1+7d,\qquad a_{44}=1+43d .$$ Step 2 - G.P. condition. $$(1+7d)^2=(1+d)(1+43d)\ \Longrightarrow\ 1+14d+49d^2=1+44d+43d^2\ \Longrightarrow\ 6d^2=30d .$$ The A.P. is non-constant, so $d\ne0$ and $d=5$. Step 3 - Sum of $20$ terms. $$S_{20}=10\left[2+19(5)\right]=10\times97=970 .$$ Answer: (B)