Question

An arithmetic progression has a second term of -14 and a sum to 21 terms of 84. Find the first term and the 21st term of this progression.​

Answer 1

${\huge{\pink{↬}}} \: \: {\huge{\underline{\boxed{\bf{\pink{Answer}}}}}}$

if the third member is a+2b

and the 22nd is a+21b

then b=(22nd member - third member)/(21–3)=(192-21)/(21–3)=171/19=9

Thus the 10th member will be a+9b=third member+7b=21+63=84

Answer 2

Given:

An arithmetic progression has a second term of -14 and a sum of 21 terms of 84. Find the first term and the 21st term of this progression.​

To find:

The first term and the 21st term of this progression.​

Solution:

Let's say,

"a" → first term of the A.P.

"d" → common difference of the A.P.

So,

The second term of the A.P. = a + d

The given arithmetic progression has a second term of -14, therefore we can form the equation as,

a + d = -14 . . . (1)

We know,

$\boxed{\bold{S_n = \frac{n}{2}[2a + (n-1)d] }}$

A sum to 21 terms of 84, therefore we can form the equation as,

$84 = \frac{21}{2}[2a + (21-1)d]$

$\implies 8 = [2a + (21-1)d]$

$\implies 8 = [2a + 20d]$

on dividing by two throughout the equation, we get

$\implies 4= a + 10d$ . . . (2)

On subtracting equations (1) and (2), we get

a + 10d = 4

a + d    = -14

-   -         +

-------------------

    9d = 18

-------------------

∴ d = 2

On substituting d = 2 in equation (1), we get

a + 2 = -14

⇒ a = -14 -2

⇒ a = -16 ← the first term of the given A.P.

We know,

$\boxed{\bold{t_n = a + (n-1)d}}$

Therefore,

The 21st term of the given A.P. is,

= $t_2_1$

= $- 16 + (21 - 1)(2)$

= $- 16 + (20)(2)$

= $- 16 + 40$

= $\bold{24}$

Thus,

$\boxed{\bold{The \:first\:term\:of\:the\:A.P.\:is\:\rightarrow \underline{-16}}}.\\\\\boxed{\bold{The \:21st\:term\:of\:the\:A.P.\:is\:\rightarrow \underline{24}}}.$

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