Sequences and Series Part 2

a1,a2,a3,.....,a10 are in AP common difference =−3
b1,b2,b3,......,b10 are in GP common ratio =2
Since, ck=ak+bk,k=1,2,3.......,10
∴c2=a2+b2=12
c3=a3+b3=13
Now, C3−C2=1
⇒(a3−a2)+(b3−b2)≠1
⇒−3+(2b2−b2)≠1
⇒b2=4
∴a2=8
So, AP is 11, 8, 5, ....
and GP is 2, 4, 8, ....
Now,

10

Σ

k=1

Ck=

10

Σ

k=1

ak+

10

Σ

k=1

bk
=(

10

2

)[22+9(−3)]+2(

210−1

2−1

)
=5(22−27)+2(1023)
=2046−25
=2021