Question

Find the mth term of an Arithmetic progression whose 12th term exceeds the 5th term by 14 and the sum of both terms is 36.​

Answer 1

To find :-

$\bigstar \sf \: find \: the \: sum \: of \: m^{th} \: term \: of \: the \: A.P$

Given :-

• $\sf {12}^{th}$ term exceeds the$\sf {5}^{th}$ term by 14 .
• Sum of $\sf {12}^{th}$ term and $\sf {5}^{th}$ term is 36 .

Formula used :-

$\underline{\boxed{\sf \:s_m \: = \dfrac{m}{2} \bigg(2a + \big(m- 1 \big)d \bigg)}}$

Solution :-

$\sf {12}^{th}$ = a + 11d .

$\sf {5}^{th}$ = a + 4d .

Where,

• a = first term of A.P
• d = common difference .

Difference b/w $\sf {12}^{th}$ term and $\sf {5}^{th}$ term = 14 .

or,

$:\implies$ a + 11d – ( a + 4d ) = 14

$:\implies$ a + 11d – a – 4d = 14

$:\implies$ 7d = 14

$:\implies\boxed{\sf d = 2 }$

Now,

sum sum of $\sf {12}^{th}$ term and $\sf {5}^{th}$term = 36 .

or,

$\:\:\:\:\:\: \:\:\: \:\:\:$ a + 11d + a + 4d = 36

$:\implies$ 2a + 15d = 36

Put the value of d = 2 we get :-

$:\implies$2a + 30 = 36

$:\implies$2a = 36 – 30

$:\implies$ 2a = 6

$:\implies\boxed{\sf a = 3 }$

Now we have the value of a and d.

$\sf \:s_m \: = \dfrac{m}{2} \bigg(2a + \big(m- 1 \big)d \bigg) \\ \\:\implies \sf \: put \: the \: value \: of \bf \: a \sf \: and \bf \: d \sf \: we \: get : \\ \\ :\implies \sf \:s_m \: = \dfrac{m}{2} \bigg(2 \times 3 + \big(m - 1 \big)2\bigg) \\ \\ :\implies \sf \:s_m \: = \dfrac{m}{2} \big(6 +2m - 2 \big) \\ \\ :\implies \sf \:s_m \: = \dfrac{m}{2} \big(4 + 2m \big) \\ \\:\implies \sf \:s_m \: = \dfrac{m}{ \cancel2} \big( \cancel2(2 + m) \big) \\ \\ :\implies \sf \:s_m \: = \dfrac{m}{2} \big(2 + m \big) \\ \\ \boxed{ \sf \:s_m \: = {m}^{2} + 2m}$