Question

how many terms of ap 3,7,11,15....gives the sum 136​

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Solution:-

Given

A.P :- 3, 7, 11, 15, ........

a = 3

d = 4

To Find :-

n = ?

Find:-

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

=) 136 = n/2 [ 2×3 + ( n-1)4 ]

=) 136 = n [ 3 + ( n-1)2 ]

=) 136 = n [ 3 + 2n - 2 ]

=) 136 = 2n² + n

=) 2n² + n - 136 = 0

=) 2n² + ( 17 - 16)n - 136 = 0

=) 2n² + 17n - 16n - 136 = 0

=) n ( 2n + 17) - 8 ( 2n + 17) = 0

=) ( 2n + 17 ) ( n -8) = 0

=) [ n = -17/2 ] and [ n = 8 ].

[ n = -17/2] Neglect because -17/2 is in Fraction.

Hence,

S8 will give the sum of 136.

Verification:-

S8 = 8/2 [ 2a + ( 8-1)d ]

=) S8 = 4 [ 2×3 + 7×4]

=) S8 = 4 [ 6 + 28]

=) S8 = 4 × 34

=) S8 = 136.

Hence Verified!