Question

Solve previous year Question of iit jee

Chapter :- sequence and series​

Answer 1

Answer:

The anwer to the question is 34

Step-by-step explanation:

⇒nth  term $a_n$ =a+(n−1)d

⇒$a_9+a_4_3$ =66

∴a+8d+a+42d=66

∴a+25d=33 ... (1)

Now, ∑  $^1^2_k=0$  $a_4_k_+_1$=416

∴13a+312d=416 (using sum of AP on the common difference parts)

∴a+24d=32 ... (2)

from (1) and (2), we get d=1 and a=8

∴∑$^1^7_k_=_1$ $a^{2}_k$ =8²+9² ⋯+24²  

 

=(1²+2²⋯+24² )−(1²+2² ⋯+7² )

=  24×25×49 ÷6 − 7×8×15÷6   (using sum of squares of n natural numbers is  

n(n+1)(2n+1)÷6

​ =4760

=140×34

Hope it is helpful

Answer 2

option (c) is the answer.