Question

an arithmetic progression consists of 37 terms. The sum of the first 3 terms of ot is 12 and the sum of the last 3 terms is 318, then find the first and last terms of the progression ​

Answer 1

Answer:

First term of the progression is 1.

Last term of the progression is 109.

Step-by-step explanation:

Given that:

An arithmetic progression consists of 37 terms.

The sum of the first 3 terms of is 12.

The sum of the last 3 terms is 318.

To Find:

The first and last terms of the progression.

Formula used:

aₙ = a + (n - 1)d

Where,

aₙ = nth terms

a = First term

d = Common difference

Let us assume:

First term be a.

Common difference be d.

Last term be l.

Sum of the first 3 terms:

⟿ a + a + d + a + 2d = 12

⟿ 3(a + d) = 12

⟿ a + d = 12/3

⟿ a + d = 4

⟿ a = 4 - d ______(i)

Sum of the last 3 terms:

⟿ l + l - d + l - 2d = 318

⟿ 3(l - d) = 318

⟿ l - d = 318/3

⟿ l - d = 106

Last term is 37.

⟿ a₃₇ - d = 106

⟿ a + 36d - d = 106

⟿ a + 35d = 106

Substituting the value of a from eqⁿ(i).

⟿ 4 - d + 35d = 106

⟿ 34d = 106 - 4

⟿ 34d = 102

⟿ d = 102/34

⟿ d = 3 ______(ii)

Finding the first term:

Now in equation (i).

⟿ a = 4 - d

Substituting the value of d from eqⁿ(ii).

⟿ a = 4 - 3

⟿ a = 1

∴ First term = 1

Finding the last term:

⟿ l = a₃₇

⟿ l = a + (37 - 1)d

Substituting the value of a and d.

⟿ l = 1 + (36 × 3)

⟿ l = 1 + 108

⟿ l = 109

∴ Last term = 109

Answer 2

$\huge \fbox \pink{answer}$

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

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

$\sf \implies 3a + d+ 2d = 12⟹3a+d+2d=12</p><p>\sf \implies 3a + 3d =12⟹3a+3d=12$

Taking 3 as common

$\sf \implies 3(a + d) = 12⟹3(a+d)=12</p><p></p><p>\sf \implies a + d = \dfrac{12}{3}⟹a+d= </p><p>3</p><p>12$

$\sf \implies a + d = 4⟹a+d=4</p><p></p><p>\sf \implies a = 4 - d⟹a=4−d$

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

Hn di bana dia..xD❤️