Question

(a2 ) × (2a23) × (4a62) find the producr plz

Answer 1

Answer:

⇒n

th

term a

n

=a+(n−1)d

⇒a

9

+a

43

=66

∴a+8d+a+42d=66

∴a+25d=33 ... (1)

Now, ∑

k=0

12

a

4k+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

∴∑

k=1

17

a

k

2

=8

2

+9

2

⋯+24

2

=(1

2

+2

2

⋯+24

2

)−(1

2

+2

2

⋯+7

2

)

=

6

24×25×49

−

6

7×8×15

(using sum of squares of n natural numbers is

6

n(n+1)(2n+1)

=4760

=140×34

Hence, the required answer is 34.