Question

14. Find the A.P., the sum of whose n terms is n(n+2).​

Answer 1

$\boxed{\textbf{\large{Step-by-step explanation:}}}$

▶As we have given , in an AP the sum of n terms is n(n+2).

Sn = n ( n + 2)

▶ In an AP (arithmetic progression) the difference between two consecutive terms is constant .

therefor, the terms are,

S1 = a = t1 = 1( 1 + 2 )

= 1 (1 + 2 )

= 3 = t1

t1 = 3

___________________

S2 = 2 ( 2 + 2 )

= 2 (4)

= 8

t2 = S2 - S1 = 8 - 3 = 5

t2 = 5

____________________

S3 = 3 ( 3 + 2 )

= 3 ( 5 )

= 15

t3 = S3 - S2 = 15 - 8 = 7

t3 = 7

_____________________

S4 = 4 ( 4 + 2 )

= 4 (6)

= 24

t4 = ( S4 - S3 )

= 24 - 15 = 9

t4 = 9

_______________________

S5 = 5 ( 5 + 2 )

= 5 ( 7)

= 35

t5 = S5 - S4

= 35 - 24

= 11

t5 = 11

______________________

therefor the AP is 3, 5, 7, 9 ,11....

Answer 2

$\mathfrak{\underline{\huge{\boxed{\red{Answer:-}}}}}$

Given :-

In A.P sum of n terms is n(n+2)

_________________

To find :-

A.P

______________________

Proof :-

1) Put n = 1

S1 = 1 ( 1+2)

S1 = 1 (3)

S1 = 3

T1 = 3

_________________________

2) Put n = 2

S2 = 2 ( 2+2)

S2 = 2 (4)

S2 = 8

T2 = S2 - S1

T2 = 8-3

T2 = 5

________________________

3) Put n = 3

S3 = 3(3+2)

S3 = 3 (5)

S3 = 15

T3 = S3 - S2

T3 = 15-8

T3 = 7

$\rule{200}{2}$

So, the A.P IS :-

3,5,7---------

Also :-

A ( First term) = 3

D(Common Difference) = 5-3 = 2