sequences and series

10
a(1+r+r²)= 56

a-1, ar-7, ar²-21 are in AP
a-1 + ar²-21 = 2(ar-7)
a(1-r)²=8

(1-r)²/(1+r+r²)= 1/7

now can you solve the rest?

a, ar, ar²,.....................ar^2n-1

sum of all terms is a(1-r²ⁿ)/(1-r)

sum of odd terms is a(1-(r²)ⁿ)/(1-r)²

the ratio is r+1=5

r=4

Q14 is very easy.. but is a bit dirty...

please dirty your hands and dont fear in handling questions :)

16

a\left(1/b+1/c\right),b(1/a+1/c),c(1/a+1/b) \text{ are in AP}

1+a\left(1/b+1/c\right),1+b(1/a+1/c),1+c(1/a+1/b) \text{ are in AP}

a\left(1/a+1/b+1/c\right),1+b(1/a+1/b+1/c),1+c(1/a+1/b+1/c) \text{ are in AP}

thus, a,b,c are in AP :)