Question

An arithmetic progression consists of 37 terms. the sum of the first 3 terms of it is 12 and the sum of it's last 3 terms is 318, then find the first and last terms of the progreesion​

Answer 1

Answer:

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⟹a+a+d+a+2d=12

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

\sf \implies 3a + 3d =12⟹3a+3d=12

Taking 3 as common

\sf \implies 3(a + d) = 12⟹3(a+d)=12

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

3

12

\sf \implies a + d = 4⟹a+d=4

\sf \implies a = 4 - d⟹a=4−d

Now

nth term = (aₙ) = a + (n - 1)d

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

\sf \implies 3(4 - d) + 105d = 318⟹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

i hope it's helpful to u. ❤

Answer 2

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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

12⟹a+a+d+a+2d=12

⟹3a+d+2d=12

⟹3a+3d=12

Taking 3 as common,

⟹3(a+d)=12

⟹a+d=12/3

⟹a+d=4

⟹a=4−d

Now,

=  (aₙ) = a + (n - 1)d 318⟹a+36d+a+35d+a+34d=318

3(4 - d) + 105d = 318⟹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

• I Hope it's Helpful My Friend.