Question

find the sum of the first 31 terms of an ap whose 4th term is 11 and 9th term is -4

Answer 1

Given :-

4th term is 11

9th term is - 4

Required to find :-

• Sum of the first 31 terms ?

Formula used :-

$\boxed{\tt{\bf{ {S}_{nth} = \dfrac{n}{2} [ 2a + ( n - 1 ) d ] }}}$

Solution :-

4th term = 11

9th term = - 4

we need to find the sum of first 31 terms .

So,

The 4th term can be represented as , a + 3d

a + 3d = 11 $\longrightarrow{\tt{ Equation - 1}}$

consider this as equation - 1

Similarly,

The 9th term can be represented as , a + 8d

a + 8d = - 4 $\longrightarrow{\tt{ Equation - 2}}$

consider this as Equation - 2

Now,

Let's solve these 2 equations simultaneously using elimination method .

subtract equation 1 & 2

$\tt{a + 3d = 11} \\ \tt{ a + 8d = - 4} \\ \underline{ \tt{( - )( - ) \: \: ( + ) \: \: }} \\ \tt{ \: \: \: \: \: - 5d = 15} \\ \\ \implies \tt - 5d = 15 \\ \\ \implies \tt d = \dfrac{ \: \: \: 15}{ - 5} \\ \\ \implies \tt d = - 3$

Hence

• Common difference ( d ) = - 3

Now,

Substitute the value of d in equation 1

a + 3d = 11

a + 3 ( - 3 ) = 11

a + ( - 9 ) = 11

a - 9 = 11

a = 11 + 9

a = 20

Hence,

• First term ( a ) = 20

Now,

Let's find the sum of first 31st terms ;

Using the formula ;

$\boxed{\tt{\bf{ {S}_{nth} = \dfrac{n}{2} [ 2a + ( n - 1 ) d ] }}}$

$\dag{\rm{ {S}_{nth} = {S}_{31} }}$

$\dag{\rm{ {S}_{31} = \dfrac{31}{2} [ 2 ( 20 ) + ( 31 - 1 ) - 3 ] }}$

$\dag{\rm{ {S}_{31} = 15.5 [ 2 ( 20 ) + ( 31 - 1 ) - 3 ] }}$

$\dag{\rm{ {S}_{31} = 15.5 [ 40 + ( 30 ) - 3 ] }}$

$\dag{\rm{ {S}_{31} = 15.5 [ 40 + ( - 90 ) ] }}$

$\dag{\rm{ {S}_{31} = 15.5 [ 40 - 90 ] }}$

$\dag{\rm{ {S}_{31} = 15.5 \times - 50 }}$

$\dag{\rm{ {S}_{31} = - 775 }}$

Therefore,

Sum of First 31st terms of the AP = - 775

Answer 2

Answer:

$\huge\boxed{s_{31} = ( - 775)}$

Step-by-step explanation:

→ Given:-

$a_4 = 11 \: \: \: and \: \: \: a_9 = ( - 4)$

→ To Find:-

$s_{31} \: (sum \: of \: 31 \: terms)$

→ Formula Applied:-

$s_n = \frac{n}{2} \times [2a + (n - 1)d]$

$a_n = a + (n - 1)d$

→ Solution:-

$a_4 = 11$

$11 = a + (4 - 1)d$

$11 = a + 3d \: \: ----(1)$

$a_9 = ( - 4)$

$( - 4) = a + (9- 1)d$

$( - 4) = a + 8d \: \: ---(2)$

Subtract Equation(1) from Equation(2):-

$11 - ( - 4) = a - a + 3d - 8d$

$15 = (- 5)d$

$d = ( - \frac{15}{5} )$

→ d=(-3)

$11 = a + 3d$

Putting value of d=(-3) :-

$11 = a + 3( - 3)$

$11 = a - 9$

$a = 11 + 9$

→ a=20

Now we have to find sum of 31 terms:

$s_{31} = \frac{31}{2} [2(20) + (31 - 1) \times ( - 3)]$

$s_{31} = \frac{31}{2} [2(20) + (30) \times ( - 3)]$

$s_{31} = \frac{31}{2} [40 - 90]$

$s_{31} = \frac{31}{2} (- 50)$

$s_{31} = 31 \times (- 25)$

$s_{31} = ( - 775)$

Hence, Sum of 31 Terms of this A.P. is (-775).