Question

how many terms of the series 51+54+57+....are required to get sum of 810

Answer 1

please mark me as Brainliest

Answer:

n=12

Step-by-step explanation:

The given series is an AP

The sum of n terms of an AP is given by  

Sn= n/2 (2a+(n-1)d)

       

a is the first term and d is the common difference.

810=n/2(102+(n-1)3)

1620=102n+3n²-3n

3n²+99n-1620=0

3(n²+33n-540)=0

n²+33n-540=0

n= -33+$\sqrt{33*33+4*1*540}$ /2

n=$\frac{-33+\sqrt{3249} }{2}$

n=$\frac{-33+57}{2}$

we have to take only positive value

n=12