Question

How many terms of the AP 3, 5, 7, … to be added to get a sum 120?​

Answer 1

Answer:

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Step-by-step explanation:

a = 3 ; d = 2 ; S n = 120

S n = n/2 [ 2a + (n-1)d ]

by solving the above eq.

n = 10 . this can be solved by factorization.

Answer 2

Given ,

First term (a) = 3

Common difference (d) = 2

Sum of n terms (Sn) = 120

We know that , the sum of first n terms AP is given by

$\boxed{ \sf{ sum = \frac{n}{2} \{2a + (n - 1)d \} }}$

Thus ,

120 = n/2 × {2(3) + (n - 1)2}

240 = n{6 + 2n - 2}

240 = n{4 + 2n}

240 = 4n + 2n²

2n² + 4n - 240 = 0

n² + 2n - 120 = 0

n² + 12n - 10n - 120 = 0

n(n + 12) - 10(n + 12) = 0

(n - 10)(n + 12) = 0

n = 10 or n = -12

Since , n = -12 is not possible

Therefore , the number of terms Is 10