Question

The first and the last term of an AP1 and 11 respectively in the sum of its terms is 36. then find the number of terms​

Answer 1

Given:-

• The first term of A.P is 1 and the last term of A.P is 11.
• The sum of its term is 36.

To Find:-

• Find the number of terms.

Solution:-

Let the common difference be d

And number of terms be x

We know that to find nth term of A.P we use the formula

→ tn = a + (n - 1)d

And,

To find nth term of an A.P

→ Sn = n/2 [ 2a + (n - 1)d ]

Now,

➣ 11 = 1 + (n - 1)d

➣ 11 - 1 = (n - 1)d

➣ ${\bf{\color{yelow}{ \: d \: = \: \frac{10}{(n \: - \: 1) \: d} }}}$ ........ ( 1 )

And,

➣ 36 = n/2 [ 2( 1 ) + (n - 1)d

➣ 72/n = 2 + (n - 1)d

➣ ${\bf{\color{yelow}{ \: d \: = \: \frac{ \: 72 \: - \: 2n }{ \: n(n \: - \: 1 ) \: } }}}$ ........ ( 2 )

Now,

Compare equation ( 1 ) = ( 2 )

${\bf{\color{yelow}{ \: \frac{10}{(n \: - \: 1) \: d} }}} = {\bf{\color{yelow}{ \: \frac{ \: 72 \: - \: 2n }{ \: n(n \: - \: 1 ) \: } }}}$

By solving we get n = 6.

Substitute n value in tn = a + (n - 1)d to get the value of d:-

➣ 11 = 1 + (6 - 1)d

➣ 11 = 1 + 5d

➣ 5d = 10

➣ d = 10/5

➣ $\boxed{\bf{\color{yelow}{ \: d \: = \: 2 \: }}}$

Verification:-

➣ 11 = 1 + (6 - 1)2

➣ 11 = 1 + 5( 2 )

➣ 11 = 1 + 10

➣ $\boxed{\bf{\color{yelow}{ \: 11 \: = \: 11 \: }}}$