# How many terms of the Ap.-17--15-13are needed to make the sum -72

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**Answer:**

**n = 6 & n = 12**

**Step-by-step explanation:**

a = -17

d = (-15)+(-17) = 2

s(n) = (-72)

s(n) = (n/2) { 2a + (n-1)d }

(-72) = (n/2) { 2(-17) + (n-1) (2) }

(-72) = (n/2) { -34 + 2n -2 }

(-72) = (n/2) { -36 + 2n }

(-72)×2 = n(2n-36)

=> 2n² - 36n = -72×2

=> n² -18n = -72

=> n² - 18n + 72 = 0

=> n² - (12+6) n + 72 = 0

=> n² - 12n -6n +72 = 0

=> n(n-12) - 6(n-12) = 0

=> (n-12) (n-6) = 0

either (n-12) =0 .....or..... (n-6) = 0

either **n**** ****=**** ****1****2**** **.......or......** ****n**** ****=**** ****6**

therefore sum of first 6 terms of AP = (-72)

also , sum of first 12 terms of AP = (-72)

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