Question

An AP consists of 50 term of which 3rd term is 12 and the last term is 106 find the 25 terms​

Answer 1

Step-by-step explanation:

this will help u out may be

Answer 2

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Given:

• Total number of terms = 50
• Third term (a3) = 12
• Last term (a50) = 106

Now,

By nth term formula,

$\boxed{\rm{\large{a_n = a + (n - 1)d}}}$

Where, an, a, n and d are the last term, first term, term number and common difference respectively.

Then,

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

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

Subtracting (1) from (2),

➛ a + 49d - (a + 2d) = 106 - 12

➛ a + 49d - a - 2d = 94

➛ 47d = 94

➛ d = 2

Then, a = 12 - 4 = 8

To Find:

Sum of 25 terms of the AP.

Then, n = 25, a = 8, d = 2.

By using formula,

$\boxed{\rm{\large{S_n = \dfrac{n}{2} \bigg (2a + (n - 1)d \bigg)}}}$

➛ S25 = 25/2 {2(8) + (25-1)2}

➛ S25 = 25/2 {16 + 24 × 2}

➛ S25 = 25/2 {64}

➛ S25 = 25 × 32

➛ S25 = 800

Hence,

The required value of the sum of 25 terms of the above AP is 800 (Ans).