Question

If the sum of first four terms of an AP is 40 and that of first 14 terms is 280 . Find the sum of its first n terms ??​

Answer 1

Step-by-step explanation:

Given : .

First Form Term A P is 40

First 14 term 280

Find : .

Sum Of n term

Solution : .

We know Sum of n term = n/2 [ 29 + ( n - 1 ) d ]

Sum Of 4 = 40 and Sum Of 14 = 280

4/2 [ 29 + ( 4 - 1 ) d ] 40 , 14/2 [ 29 + ( 14 - 1 ) d ] = 280

( 29 + 3d ) = 40/2 , 7 ( 29 + 13d ) = 280

29 + 3d = 20 ( i )

29 + 13d = 40 ( ii )

From Equation ( i ) and ( ii )

29 + 3d = 20

29 + 13d = 40

- 10d = - 20

d = - 20/-10

d = 2

Putting value d = 2

From ( i ) we get a = 7

Sum Of n term = n/2 [ 2 ( 7 ) + ( n - 1 ) 2 ]

Sum of n term = n/2 2 [ 7 + n - 1 ]

Sum of n term = n ( n + 6 )

Sum of n term n ( n + 6 ) or n^2 + 6 n

Answer 2

✔️ Solution-

We know that,

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

Therefore, S4 = 40

$= > \: \frac{4}{2}[2a + (4 - 1)d] = 40 \\ (given)$

$= > \: 2a + 3d = 20...(1)$

$3_{14} = 280$

$= > \: \frac{14}{2}[2a + (14 - 1)d] = 280 \\ (given)$

$= > \: 2a + 13d = 40...(2)$

After solving (1) and (2), we get

a = 7

d = 2

Therefore,

$S_{n} = \frac{n}{2}[2 \times 7 + (n - 1)2]$

$= > \: S_{n} = n(7 + n - 1)$

$= > \:S _{n} = n(n + 6)$