Question

In an A.P. of 50 terms, the sum of first 10 terms is 210 and the sum of its last 15 terms is 2565. Find the A.P.

Answer 1

■ Let ' a ' be the first term and ' d ' be the common difference of the required A.P.

nth term of A.P. =》

An = a + ( n - 1 ) d

■ And the sum of first n terms of A.P. =》

Sn = n/2 [ 2a + ( n - 1 ) d ]

Sum of first 10 terms ,

S10 =》 210 = 10 / 2 [ 2 * a + ( 10 - 1 ) d

=》210 = 5 [ 2a + 9d ]

=》42 = 2a + 9d

2a + 9d = 42 ................( 1 )

■ 15th term from the last = ( 50 - 15 + 1 )

=》36th term

An = a + ( n - 1 ) d

An = a + ( n - 1 ) d

a36 = a + 35d

■ Sum of last 15 terms =》

s15 =》2565 = 15 / 2 [ 2a + ( 15 - 1 ) d ]

Put the value of a36 at " a "

2565 = 15 / 2 [ 2 ( a + 35d ) + 14d ]

2565 = 15 [ a + 35d + 7d ]

a + 42d = 171 ....................( 2 )

■ Subtracting eq. ( 2 ) from eq. ( 1 ) , we get ,

First multiply eq. ( 2 ) by 2 =》

2a + 84d = 342

On subtracting ,

9d - 84d = 42 - 342

75d = 300

d = 300 / 75

d = 4

■ Put the value of d in eq. ( 2 )

a + 42d = 171

42 * 4 + a = 171

168 + a = 171

a = 171 - 168

a = 3

$\textbf{So , the A.P. formed is 3 , 7 , 11 , 15 .......... 199}$

Thanks

Answer 2

$\underline{ \underline{\bf{Answer}}} : - \\ \implies \: 3, \: 7 \:, 11 \: ,15, \: ..........,199 \\ \\ \underline{\underline{ \bf{Step - by - step \: explanation \: }}} : -$

According to the question:-

$\bf{sum \: of \: first \: 10 \: terms \:( s_{10}) = 210} \\ 210 = \frac{10}{2} \bigg (2a + (101)d \bigg) \: \\ \\ 2a + 9d = 42 \: .........(1)\\ \\ \bf{sum \: of \: last \: 15 \: terms \: ( s_{15})= 2565} \\ \\ s_{50} -s_{35} = 2565 \\ \\ 2565 = \frac{50}{2} \bigg(2a + (50 - 1)d \bigg) - \frac{35}{2} \bigg(2a + (35 - 1)d \bigg) \\ \\ 2565 = 25(2a + 49d) - 35(a + 17d) \\ \\ 2565 = 50a + 1225d - 35a - 595d \\ \\ after \: solving \: this \: \\ \\ a + 42d = 171 \: ...........(2) \\ \\ from \: eq(1) \: and \: (2) \\ \\eq (1) \times 42 - \: eq (2) \times 9 \\ \\ we \: get \: \\ \\ a = 3 \: d = 4$

Hence required AP is →

3,7,11,15,....,199