Question

How many terms of an A.P 42,36,30... amount to 150

Answer 1

Given

42, 36, 30 ...... form an AP

To find

The number of terms that sum up to 150

FORMULA USED

Sₙ= n/2 [ 2a + ( n - 1 ) d ]

Where

Sₙ = Total Sum

a = First term

n = Number of terms

d = Common difference

$\implies 150 = \dfrac{n}2}[2*42 + (n-1)-6]$

$\implies 300=n[84-6n+6]$

$\implies 300 = n(90-6n)$

$\implies 300 = -6n^{2} +90n$

$\implies 6n^{2} -90n+300=0$

$\implies 2n^{2} -30n+100=0$

$\implies 2n^{2} -10n-20n+100=0$

$\implies 2n(n-5)-20(n-5)=0$

$\implies (2n-20)(n-5) = 0$

Now,

$2n-20=0\\\implies 2n=20\\\implies n = 10$

$n-5=0\\\implies n = 5$

Therefore, there are 2 terms for n which are possible : 5 and 10