Question

how many terms of arithmetic sequence 99,97,95.... must be added to get 900?​

Answer 1

Step-by-step explanation:

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

Here Sn=900

a=99

d=-2

900=n*{198+(n-1)*(-2)}/2

Or 1800=198n+(n^2-n)(-2)

Or 900=99n-n^2+n

Or n^2–100n+900=0

Or (n-10)(n-90)=0

So n=10,90

It indicates that sum of 11th term to 90th is zero.

We must add either first 10 terms or first 90 terms to get 900

Answer 2

Answer:

Thus sum of first 10 terms and of 90 terms =900

Step-by-step explanation:

99,97,95....

here a=99 d=97-99=-2

Let sum of n terms=900

we know that

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

900=n/2 { 2*99+(n-1)(-2) }

900=n/2 ( 198-2n+2)

900=n/2( 200-2n)

900=2*n/2*(100-n)

900=100n-n²

n²-100n+900=0

n²-90n-10n+900=0

n(n-90)-10(n-90)=0

(n-90) (n-10)=0

So n=10 or 90

Thus sum of first 10 terms and of 90 terms =900

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How this can happen:

First 10 terms are positive and total=900

then 50th term =1 is the last positive term

S50=2500

Then 51st to 90th terms total= -1600

Thus sum of 90 terms =2500-1600=900

NB: Finding the values of S10,S50 and S90

I leave for students

Thanks for asking a nice question