Question

please answer these question ​

Answer 1

Step-by-step explanation:

nth term of an Ap = 6n+3

on putting n=1 we get first term i.e (a)

Tn= 6n+3

T1= 6×1+3

T1= 9

a=9

if nth term of an Ap be in equation of (an+b)

then cofficient of n be common difference.

here,

tn= 6n+3

cd= 6

Sn=n/2(a+l) [l=last term]

765= n/2(9+6n+3)

765×2= n(12+6n)

1530= n×6(2+n)

255= n(2+n)

255= 2n+n²

n²+2n-255=0

n²+17n-15n-255=0

n(n+17)-15(n+17)=0

(n-15)(n+17)=0

n-15= 0

n=15

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