Question

The sum of first seven terms of an A.P. is 49 and the sum of first
eleven terms of the same A.P. is 121. Find the sum of 25 terms.

Answer 1

Answer:

sum =625

Step-by-step explanation:

step by step explanation given above

Answer 2

Given : -

Sum of first seven terms of an AP = 49

Sum of first eleven terms of an AP = 121

Required to find : -

• Sum of first 25 terms

Formula used : -

To find the sum of nth terms of any given arithmetic progression / sequence is ;

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

Solution : -

Sum of first seven terms of an AP = 49

Sum of first eleven terms of an AP = 121

we need to find the sum of first 25 terms ?

So,

Sum of first 7 terms = 49

This implies ;

$\sf S_{nth} = S_{7} \\ \\ \sf S_7 = \dfrac{7}{2} [ 2a + ( 7 - 1 )d ] \\ \\ \sf S_{7} = \frac{7}{2} [ 2a + ( 6 ) d ] \\ \\ \tt \because S_7 = 49 \\ \\ \rm 49 \times 2 = 7(2a + 6d) \\ \\ \rm 98 = 7(2a + 6d) \\ \\ \rm \frac{98}{7} = 2a + 6d \\ \\ \rm 14 = 2a + 6d \\ \\ \rm 2a + 6d = 14 \longrightarrow equation - 1$

Consider this as equation 1

Similarly,

Sum of first eleven terms of the AP = 121

So,

This implies ;

$\sf S_{nth} = S_{11} \\ \\ \sf S_11 = \dfrac{11}{2} [ 2a + ( 11 - 1 )d ] \\ \\ \sf S_{11} = \frac{11}{2} [ 2a + ( 10 ) d ] \\ \\ \tt \because S_{11} = 121 \\ \\ \rm 121\times 2 = 11(2a + 10d) \\ \\ \rm 121 \times 2 = 11(2a + 10d) \\ \\ \rm \frac{121 \times 2}{11} = 2a + 10d \\ \\ \rm 22= 2a + 10d \\ \\ \rm 2a + 10d = 22 \longrightarrow equation - 2$

Consider this as equation - 2

Now,

We need to solve these two equations simultaneously .

In order to solve them let's use Elimination method because we can eliminate an variable to simply our calculations .

Subtract equation 1 from equation 2

$\sf 2a + 10d = 22 \\ \sf 2a + \: \: 6d = 14 \\ \underline{( - )( - ) \: \: \: \: ( - ) \: \: \: } \\ \underline{ \: \: \: \: \: \: \: \sf \: \: \: \: \: 4d = 8 \: \: \: } \\ \\ \implies \tt 4d = 8 \\ \\ \tt \implies \tt d = \frac{8}{4} \\ \\ \implies \tt d = 2$

Hence,

• Common difference ( d ) = 2

Substitute the value of d in Equation 1

2a + 6d = 14

2a + 6 ( 2 ) = 14

2a + 12 = 14

2a = 14 - 12

2a = 2

a = 2/2

a = 1

Hence,

• First term ( a ) = 1

Now,

Let's find the sum of first 25 terms ;

$\tt S_{nth} = S_{25} \\ \\ \tt S_{25} = \dfrac{25}{2} [ 2 ( 1 ) + ( 25 - 1 ) 2 ] \\ \\ \tt S_{25} = \dfrac{25}{2} [ 2 + ( 24 )2 ] \\ \\ \tt S_{25} = \dfrac{25}{2} [ 2 + 48 ] \\ \\ \tt S_{25} = \dfrac{25}{ 2 } [ 50 ] \\ \\ \tt S_{25} = \dfrac{25}{ 2} \times 50 \\ \\ \tt S_{25} = 25 \times 25 \\ \\ \tt S_{25} = 625$

Therefore,

Sum of first 25 terms of the AP = 625