Let a1 , a2 , a3 , … be in an arithmetic progression of positive terms. Let Ak = a12 - a22 + a32 - a42 + … + a2k - 12 - a2k2 . 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 _______ .

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Solution

Q.
Let $a_{1}, a_{2}, a_{3}, \dots$ be in an arithmetic progression of positive terms.

Let $A_{k} = a_{1}^{2} - a_{2}^{2} + a_{3}^{2} - a_{4}^{2} + \dots + a_{2 k - 1}^{2} - a_{2 k}^{2} .$

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 _______ . [2024]

Ans.
(910)

$a_{1}, a_{2}, a_{3}, \dots$ is an A.P.

Let $d$ be the common difference

$∴$    $a_{2} - a_{1} = a_{3} - a_{2} = \dots = a_{2 k} - a_{2 k - 1} = d$

Now, $a_{1}^{2} - a_{2}^{2} + a_{3}^{2} - a_{4}^{2} + \dots + a_{2 k - 1}^{2} - a_{2 k}^{2}$

$= (a_{1} - a_{2}) (a_{1} + a_{2}) + \dots + (a_{2 k - 1} - a_{2 k}) (a_{2 k - 1} + a_{2 k})$

$= - d [a_{1} + a_{2} + \dots + a_{2 k - 1} + a_{2 k}]$

$= - d \cdot [\frac{2 k}{2} [a_{1} + a_{2 k}]] = - d k (a_{1} + a_{2 k})$

$∴$ $A_{k} = - d k (a_{1} + a_{2 k})$

So,   $A_{3} = - 3 d (a_{1} + a_{6}) = - 153$

$\Rightarrow - 3 d (a_{1} + a_{1} + 5 d) = - 153$

$\Rightarrow - 3 d (2 a_{1} + 5 d) = - 153 \Rightarrow 2 a_{1} + 5 d = \frac{51}{d}$ ...(i)

Similarly $A_{5} = - 435$

$\Rightarrow - 5 d (2 a_{1} + 9 d) = - 435 \Rightarrow 2 a_{1} + 9 d = \frac{87}{d}$ ...(ii)

Solving (i) and (ii) we get $4 d = \frac{36}{d} \Rightarrow d^{2} = 9 \Rightarrow d = 3$ [ $∵$ Given A.P. is a progression of positive terms]

$\Rightarrow a_{1} = 1$

Now, $a_{17} = a_{1} + 16 d = 1 + 48 = 49$

and $A_{7} = - 3 \times 7 (a_{1} + a_{14})$

$= - 21 (1 + 1 + 13 \times 3) = - 21 (41) = - 861$

$∴$    $a_{17} - A_{7} = 49 + 861 = 910$

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