Question

Find the sum of first ten term of A.P whose 4th term -15 and 9th term is-30​

Answer 1

Answer:

a4 = -15

a9 = -30

a+3d=-15

a+8d = -30

-5d = 15

d = -3

a-9 = -15

a =-6

s10 = 5(-12+(-27))

s10 = 5(-39)

s10 = -195

Answer 2

Answer:

THE ANSWER IS

Step-by-step explanation:

Since, the total number of terms(n)=11 (n)=11 [odd]

∴∴ Middle most term=(n+1)2th term=(11+12)=(n+1)2th term=(11+12)th term = 6th term

Given that, a6=30a6=30

⇒a+(6−1)d=30⇒a+(6-1)d=30

⇒a+5d=30 ⇒a+5d=30…(i)

∵∵ Sum of n term of an AP, Sn=n2[2a+(n−1)d]Sn=n2[2a+(n-1)d]

∴S11=112[2a+(11−1)d]∴S11=112[2a+(11-1)d]

 =112[2a+10d]=11(a+5d) =112[2a+10d]=11(a+5d)[from Eq. (i)]

 =11×30=330