Question

★$Question:-$

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

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Answer 1

Solution -

It is given that the arithmetic progression consists of 37 terms . Let us assume that the first term is a and the common difference is d. Then the last term becomes ( a + 36d).

The sum of the first three terms is 12.

So

a + a + d + a + 2d = 12

> 3a + 3d = 12

> a + d = 4

Now

The sum of the last three terms is 318

> a + 34d + a + 35d + a + 36d = 318

> 3a + 105d = 318

> a + 35d = 106

Subtracting this from the first equation

> 34d = 102

> d = 3

> a = 1

Last term > a + 36d = 1 + 108 = 109

Answer : The first and last terms of the ap are 1 and 109 respectively .

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Answer 2

Question:

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

Given:

• An Arithmetic progression consists of 37 terms.
• The sum of first three terms = 12
• The sum of last three terms = 318

To Find:

The first and Last term of Arithmetic progression.

Solution:

Let us assume that a is the first term and d is the common difference of the Arithmetic progression. Then,

the last term of Arithmetic progression becomes (a+36d)

Now,

As we are given that sum of first three term is 12. So,

$\: \dashrightarrow\mathsf{(a)+(a+d)+(a+2d)=12}$

$\dashrightarrow\mathsf{a+a+d+a+2d=12}$

$\dashrightarrow\mathsf{a+a+a+2d+d=12}$

$\dashrightarrow\mathsf{3a+3d=12}$

$\dashrightarrow\mathsf{3(a+d)=12}$

$\dashrightarrow\mathsf{a+d=\large\frac{12}{3}}$

$\dashrightarrow\mathsf{a+d=4} -(1)$

Now,

It is also given that sum of last three terms of Arithmetic progression is 318. So,

$\\ \dashrightarrow\mathsf{(a+34d)+(a+35d)+(a+36d)=318}$

$\dashrightarrow\mathsf{a+34d+a+35d+a+36d=318}$

$\dashrightarrow\mathsf{3a+105d=318}$

$\dashrightarrow\mathsf{3(a+35d)=318}$

$\dashrightarrow\mathsf{a+35d=\large\frac{318}{3}}$

$\dashrightarrow\mathsf{a+35d=106} -(2)$

Subtracting eqn (1) from (2). we get,

$\dashrightarrow\mathsf{34d=102}$

$\\ \dashrightarrow\mathsf{d=\large\frac{102}{34}}$

$ㅤㅤ \\ \dashrightarrow\mathsf{d=3}$

Putting value of d in eqn (1). We get,

$ㅤㅤ\dashrightarrow\mathsf{a+3 =4}$

$ㅤㅤ\dashrightarrow\mathsf{a=1}$

Lets find last term of Arithmetic progression,

$ㅤ\dashrightarrow\mathsf{a+36d}$

$ㅤ\dashrightarrow\mathsf{1+108}$

$ㅤ\dashrightarrow\mathsf{109}$

$\therefore$ first and last term of an AP are 1 and 109 respectively.