Question

.
Find the sum of first 16 terms of an Arithmetic Progression whose
4th and 9th terms are - 15 and - 30 respectively.

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Answer 1

Answer:

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Step-by-step explanation:

a4= a+(n-1)d

=> a+(4–1)d= -15

=> a+3d= -15

=> a= -15-3d → eq.1

a9= a+(9–1)d

=>a+8d= -30

=> a= -30-8d →eq. 2

From equation 1 and 2, we get

-15-3d= -30-8d

=> -3d+8d= -30+15

=> 5d= -15

=> d= -3

Putting d= -3 in eq. 1,

a= -15–3(-3)= -15+9= -6

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

=> S15= 15/2 [2(-6)+14(-3)]= 15/2(-54)= -135

Therefore, the sum of first 15 terms of the given AP is -135.

Answer 2

See the above given photo for answer and proper explanation