Find the sum of first 15 terms of an AP whose 4th and 9th t terms are -15 and -30 respectively.

Sum of the first 15 terms = -405

Step-by-step explanation:

a + 3d = -15 (1)

a + 8d = -30 (2)

Where,

a = first term

d = common difference

n = number of terms

Subtract (1) from (1)

8d - 3d = -30 - (-15)

5d = -30 + 15

5d = -15

d = -15/5

= -3

d = -3

Substitute d = -3 into (1)

a + 3d = -15

a + 3(-3) = -15

a - 9 = -15

a = -15 + 9

a = -6

Sum of the first 15 terms

S = n/2[2a + (n − 1) × d]

= 15/2 {2×-6 + (15-1)-3}

= 7.5{-12 + (14)-3}

= 7.5{ -12 - 42}

= 7.5{-54}

= -405

Sum of the first 15 terms = -405

this is a very good question

step-by-step explanation:

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