Question

If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of first n terms.

ALIGARH MUSLIM UNIVERSITY BOARD OF SECONDARY AND SENIOR SECONDARY EDUCATION, SAMPLE PAPER

Answer 1

HEYA!!!!!

Given,

Sum of first seven terms = 49

sum of seventeen terms = 289

then, Sum of first n terms = ?

49 = n/2 ( 2a + ( n-1) d )

49 = 7/2 ( 2a + 6d)

7/2(2a + 6d)= 49

2a + 6d = 49 × 2/7

2a + 6d = 14 -----------(1)

S17 = 289

n/2 ( 2a + ( n-1) d ) = 289

17/2 ( 2a + 16d) = 289

2a +16d = 289 × 2/17

2a + 16d = 34 ----------(2)

from (1) & (2)

2a + 6d = 14

(+)-2a (+) - 16d = 34
----------------------
- 10 d = -20

d = 20/10

d = 2

Substitute d=2 in eq - (1)

2a + 6 ( 2) = 14

2a = 14-12

2a = 2

a = 2/2

a = 1

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

= n/2 ( 2 (1) + ( n-1) 2 )

= n/2 ( 2+2n-2 )

= n/2 × 2n

= n × n

Sn. = ${n}^{2}$

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HOPE THIS HELPS U.

Answer 2

HELLO DEAR,

given that:-


$s _{7} = 49 \\ = > 49 = \frac{7}{2} (2a + (7 - 1)d) \\ = > \frac{49}{7} = \frac{1}{2} (2a + 6d) \\ = > 7 = a + 3d...... .......(1)$
AND
$s_{17} = 289 \\ = > 289 = \frac{17}{2} (2a + (17 - 1)d) \\ = > 289 = \frac{17}{2} (2a + 16d) \\ = > \frac{289}{17} = \frac{1}{2} (2a + 16d) \\ = > 17 = a + 8d.......(2)$

SUBTRACTING from (1)&(2)

we get,

a+3d=7
a+8d=17
(-) (-) (-)
=============
-5d= -10

d=2 put in - - (1)

We get,

7=a+3*2

7=a+6

a=7-6

a=1

$s_{n } = \frac{n}{2} (2a + (n - 1)d) \\ = > s_{n} = \frac{n}{2} (2 \times 1 + 2n - 2) \\ = > s_{n} = n \times n \\ = > s_{n} = {n}^{2}$
I HOPE ITS HELP YOU DEAR,
THANKS