JEE Main 2024 (Online) 5th April Morning ShiftQuestion 1
Let $a_1,a_2,a_3,\ldots$ be in an arithmetic progression of positive terms.
Let $A_k=a_1^{2}-a_2^{2}+a_3^{2}-a_4^{2}+\ldots+a_{2k-1}^{2}-a_{2k}^{2}$.
If $A_3=-153$, $A_5=-435$ and $a_1^{2}+a_2^{2}+a_3^{2}=66$, then $a_{17}-A_7$ is equal to $\_\_\_\_\_\_\_\_$ .
Sequence and Series · Arithmetic Progression
Answer$910$
Solution
Formulas used. (1) $x^2-y^2=(x-y)(x+y)$(2) $S_n=\frac n2\left[2a+(n-1)d\right]$Step 1 - Simplify $A_k$. Each pair gives $a_{2j-1}^2-a_{2j}^2=(a_{2j-1}-a_{2j})(a_{2j-1}+a_{2j})=-d\,(a_{2j-1}+a_{2j})$. Adding the $k$ pairs: $$A_k=-d\,S_{2k}.$$ Step 2 - The two conditions. $$A_3=-d\cdot3(2a+5d)=-153\ \Longrightarrow\ d(2a+5d)=51,$$ $$A_5=-d\cdot5(2a+9d)=-435\ \Longrightarrow\ d(2a+9d)=87 .$$ Step 3 - Solve. Subtracting, $4d^2=36$, so $d=\pm3$. With $d=-3$, $2a-15=-17$ gives $a=-1$, not positive. So $d=3$ and $2a+15=17$, $a=1$. Check: $1^2+4^2+7^2=66$. Step 4 - Required value. $$a_{17}=1+16(3)=49,\qquad A_7=-3\,S_{14}=-3\cdot7(2+39)=-861 .$$ $$a_{17}-A_7=49+861=910 .$$ Answer: 910