Question

how many terms of an ap 17, 15, 13 ... must be added to get the sum 72 ?

Answer 1

We have been given the AP:-

17, 15, 15...

The first term is 17.

So, a = 17

Firstly, find the value of 'd':-

d = 15 - 17 = -2

So, the common difference is -2.

Now, the formula for sum of terms in an AP is:-

$S_{n}$ = $\frac{n}{2}$ x [ 2a + (n - 1)d]

Now, substitute the values in the formula:-

72 =  $\frac{n}{2}$ x [ 2(17) + (n - 1)-2]

72 = $\frac{34n-2n^{2}+2n}{2}$

Add the like terms:-

72 = $\frac{36n-2n^{2}}{2}$

Cross multiplication:-

144 = 36n - $2n^{2}$

Transposing:-

$2n^{2}$ - 36n + 144 = 0

$n^{2}$ - 18n + 72 = 0

Middle Term Splitting:-

$n^{2}$ - 12n - 6n + 72 = 0

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

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

So, n - 12 = 0 or n - 6 = 0

That means, n = 12 or 6

So, we have to add 12 consecutive terms of AP to get 72 or add 6 terms of the AP to get 72.

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Answer 2

Answer:

the above attachment will help

Step-by-step explanation:

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