Question

An AP consists of 50 term of which 3rd term is 12 and the last term is 106 find the 69 terms. ​

Answer 1

$\huge\underline\mathbb\purple{ANSWER}$

144

$\huge\underline\mathbb\green{EXPLANATION}$

Let a , d are first term and common

difference of an A.P

nth term = Last term = a + (n - 1)d

an = a + (n - 1 )d

Now,

It is given that,

Third term = 12

a + 2d = 12 ------(1)

Last term = 106

a + 49d = 106 ---(2) =

Subtract (1) from (2), we get

47d = 94

d = 2

Substitute d value in equation (1),

We get

a + 2 x 2 = 12

a = 12 - 4

a = 8

69th term = a + 69d

a69 = 8 + 68 x 2

= 8 + 136

= 144

Hope it helps ✌

Answer 2

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