Question

How many terms of the arithmetic sequence 99,97,95 must be added to get 900? Can anyone answer fast ​

Answer 1

Answer:

If we take 10 then after ten terms its sum would become 900. If we take 90, then after 10 terms its sum would be 900 and then it will increase until a certain point and then again the sum will start decreasing because of negative values which will continue till 90th term making the sum 900 again.

Answer 2

Step-by-step explanation:

Given: $99, \: 97, \: 95 \: ...$

To find: How many terms of the arithmetic sequence 99,97,95 must be added to get 900?

Solution:

Tip: Sum of n terms of AP

$\bf \red{S_n = \frac{n}{2} [2a + (n - 1)d]}$

Step 1: Write the given values.

$a = 99$

$d = - 2$

$S_n = 900$

Step 2: Put values in the formula and find n.

$900 = \frac{n}{2} [2 \times 99 + (n - 1)( - 2)]$

$900 = n [ 99 - (n - 1)]$

$900 = 99n - {n}^{2} + n$

${n}^{2} - 100n + 900 = 0$

${n}^{2} - 90n - 10n + 900 = 0$

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

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

$\bf n = 10,\: 90$

Verification:

Case 1: When $n = 10$

$S_{10} = \frac{10}{2} [2 \times 99 - 2(10 - 1)]$

$S_{10} = 10(99 - 9)$

$\bf \red{S_{10} = 900}$

Case 2: When $n = 90$

$S_{90} = \frac{90}{2} [2 \times 99 - 2(90 - 1)]$

$S_{90} = 90(99 - 89)$

$S_{90} = 90 \times 10$

$\bf \green{S_{90} = 900}$

Final answer:

Number of terms n=10, 90 in AP 99,97,95... to get a sum of 900.

Hope it helps you.

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