Question

How man
many terms of the A.P.2,46
must be taken so that sum
is 10100 ?​

Answer 1

Answer:

100

Step-by-step explanation:

First term = a = 2

Com. differ. = d = 4 - 2 = 2

Let the required number of terms be 'n'.

Using S = (n/2) {2a + (n - 1)d}

=> 10100 = (n/2) {2(2) + (n-1)2}

=> 20200 = n(4 + 2n - 2)

=> 20200 = 2n² + 2n

=> 10100 = n² + n

=> n² + n - 10100 = 0

=> n² + 101n - 100n - 10100 = 0

=> n(n + 101) - 100(n + 101) = 0

=> (n + 101)(n - 100) = 0

=> n = 100

Required number of terms = 100

Answer 2

Given ap=2,4,6.......
a=2, d=2
Sn=n(a+l)/2 where sis the 1st term, l is nth term
an=a+(n-1)d=2+(n-1)2=2n
Sn=n(2+2n)/2=n(1+n)
But given Sn=10100
So 10100=n(1+n)
n^2+n-10100=0
n^2+101n-100n-10100=0
n(n+101)-100(n+101)=0
(n+101) (n-100)=0
(n+101)=0 or (n-100)
n=-101 or n=100
n=-101 is not possible so n=100
100 terms must be taken to get the sum ofAP(2,4,6,..........) as 10100