Question

Find the least value of n which sum of the series 12+20...........to n terms is greater than 1000?​

Answer 1

Answer:

S=1−(1/2)1=2,

Sn=1−(1/2)1−(1/2)n=2−2n−11

S−Sn=2n−11<10001 or 2n−1≥1000

Now 210=32×32=1024

∴n−1≥10 or n≥11

Hence the least value is 11.

Answer 2

Step-by-step explanation:

here's the ans

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