Question

*Plz answer this question as fast as possible*​

Answer 1

Answer:

Option(B)

Step-by-step explanation:

First term = a and last term = l.

Let n be number of terms of an AP.

Sn = (n/2)[2a + (n - 1) * d]

    = (n/2)[a + a + (n - 1) * d]

    = (n/2)[a + l]{l = a + (n - 1) * d}

Hence, Sum of n terms = (n/2)[a + l]

S = (n/2)[a + l]

=> 2S = n[a + l]       --------- (1)

We have,

l = a + (n - 1) * d

=> d = (l - a)/(n - 1)

Common difference is d = l^2 - a^2/k - (l + a)

=> (l - a)/(n - 1) = l² - a²/k - (l + a)

=> (l - a)/(n - 1) = (l + a)(l - a)/k - (l + a)

=> 1/(n - 1) = l + a/k - (l + a)

=> k - (l + a) = (l + a)(n - 1)

=> k = (l + a)(n - 1) + (l + a)

=> k = l + a[n - 1 + 1]

=> k = l + a * n

=> k = 2S{From (1)]

Hence, the value of k = 2S.

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