Question

An Arithmetic progression consists of 37 terms.The sum of first 3 terms is 12 and the sum of last 3 terms is 318,then find the first and last term of the AP​

Answer 1

Answer:-

Given:-

Number of terms in an AP (n) = 37

Sum of first three terms = 12

First three terms of an AP are a , a + d , a + 2d.

So,

⟹ a + a + d + a + 2d = 12

⟹ 3a + 3d = 12

⟹ 3(a + d) = 12

⟹ a + d = 12/3

⟹ a = 4 - d -- equation (1).

Also given that,

Sum of last three terms (i.e., 37th , 36th , 35th) is 318.

We know,

nth term of an AP (aₙ) = a + (n - 1)d

According to the above condition,

⟹ a + (37 - 1)d + a + (36 - 1)d + a + (35 - 1)d = 318

⟹ 3a + 36d + 35d + 34d = 318

Substitute the value of a from equation (1).

⟹ 3(4 - d) + 105d = 318

⟹ 12 - 3d + 105d = 318

⟹ 102d = 318 - 12

⟹ d = 306/102

⟹ d = 3

Substitute the value of d in equation (1).

⟹ a = 4 - 3

⟹ a = 1

Now,

last term is the 37th term.

⟹ a₃₇ = a + 36d

⟹ a₃₇ = 1 + 36(3)

⟹ a₃₇ = 1 + 108

⟹ a₃₇ = 109

∴

• The first term is 1.

• The last term is 109.

Answer 2

Given :-

In AP consists of 37 terms. The sum of first 3 terms is 12 and the sum of last 3 terms is 318

To find :-

1st and last term of AP

Solution :-

Here the first 3 terms are

a

a + d

a + 2d

$\sf \implies a + a + d + a + 2d = 12$

$\sf \implies 3a + d+ 2d = 12$

$\sf \implies 3a + 3d =12$

Taking 3 as common

$\sf \implies 3(a + d) = 12$

$\sf \implies a + d = \dfrac{12}{3}$

$\sf \implies a + d = 4$

$\sf \implies a = 4 - d$

Now

nth term =  (aₙ) = a + (n - 1)d

$\sf \implies a + 36d + a + 35d + a + 34d = 318$

$\sf \implies 3(4 - d) + 105d = 318$

⟹ 12 - 3d + 105d = 318

⟹ 105d - 3d = 318 - 12

⟹ 102d = 318 -12

⟹ 102d = 306

⟹ d = 306/102

⟹ d = 3

⟹a = 4 - 3

⟹ a = 1

Now

⟹ a₃₇ = a + 36d

Putting d as 3

⟹ a₃₇ = 1 + 36(3)  

⟹ a₃₇ = 1 + 108  = 109