Question

-10, –12, -14, -16,... upto 15 terms,Find the sum of the first n terms of the following AP.as asked for.

Answer 1

a = - 10

d = - 12

n = 15

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

= 15 /2 [ 2 (-10) + 14 (-2)]

= 15 /2 ( - 20 - 28)

= 15 × ( - 48)/ 2

= 15 ( - 24)

= - 360

Answer 2

Given A.P :

-10 , -12 , -14 , - 16 , ...., 15 terms

first term (a) = -10

Common difference (d)=a2 - a1

=> d = -12 - ( -10 )

=> d = -12 + 10

=> d = -2

i ) Sum of n terms ( Sn )

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

Here ,

a = -10 , d = -2 , n = 15

=> S15 = ( 15/2 )[2(-10)+(15-1)(-2)]

= (15/2)[ -20 + 14 (-2)]

= ( 15/2 )[ -20 - 28 ]

= ( 15/2 ) ( -48 )

= 15 × ( -24 )

= -360

S15 = -360

ii ) Sum of n terms ( Sn )

= n/2 [ 2×(-10 ) + ( n - 1 )(-2)]

= n/2[ -20 - 2n + 2 ]

= n/2[ -18 - 2n ]

= 2n /2 [ -9 - n ]

= ( -n )( 9 + n )

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