Find the sum:34 + 32 + 30 + . . . + 10

1.	Find the sum:34 + 32 + 30 + . . . + 10
Answer» We know that the sum of terms for an A.P is given by S_n = \(\frac{n}{2}\)[2a + (n − 1)d] Where; a = first term for the given A.P. d = common difference of the given A.P. n = number of terms Or S_n = \(\frac{n}{2}\)[a + l] Where; a = first term for the given A.P. ;l = last term for the given A.P Given series, 34 + 32 + 30 + . . . + 10 which is an A.P Where, a = 34 ,d = 32 – 34 = -2 and last term (a_n= l) = 10 We know that, a_n = a + (n – 1)d So, 10 = 34 + (n – 1)(-2) 10 = 34 – 2n + 2 n = \(\frac{(36 \,–\, 10)}{2}\) = 13 Now, for the sum of these 13 terms S₁₃= \(\frac{13}{2}\) [34 + 10] = \(\frac{13(44)}{2}\) = 13 x 22 = 286 Hence, the sum of terms of the given series is 286.

Answer»

We know that the sum of terms for an A.P is given by

S_n = \(\frac{n}{2}\)[2a + (n − 1)d]

Where; a = first term for the given A.P. d = common difference of the given A.P. n = number of terms

Or S_n = \(\frac{n}{2}\)[a + l]

Where; a = first term for the given A.P. ;l = last term for the given A.P

Given series, 34 + 32 + 30 + . . . + 10 which is an A.P

Where, a = 34 ,d = 32 – 34 = -2 and last term (a_n= l) = 10

We know that, a_n = a + (n – 1)d

So,

10 = 34 + (n – 1)(-2)

10 = 34 – 2n + 2

n = \(\frac{(36 \,–\, 10)}{2}\) = 13

Now, for the sum of these 13 terms

S₁₃= \(\frac{13}{2}\) [34 + 10]

= \(\frac{13(44)}{2}\)

= 13 x 22

= 286

Hence, the sum of terms of the given series is 286.

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