What is the sum of the arithmetic sequence 6, 14, 22 ..., if there are 26 terms?
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Given (A.P)
first term, a1 = 6
number of terms, n = 26
common difference, d = 8
To find: sum of 26 terms of A.P
Sum = n/2 * (a1+an)
= n/2 * (a1 + a1 + (n-1)*d)... [an = a1+(n-1)*d]
= 26/2 * [6 + 6 + (26-1)*8]
= 13*[12 + 25*8] = 13*[12+200] = 2756
Hence sum of 26 terms of given arithmetic sequence is 2756.
Hope it helps.
first term, a1 = 6
number of terms, n = 26
common difference, d = 8
To find: sum of 26 terms of A.P
Sum = n/2 * (a1+an)
= n/2 * (a1 + a1 + (n-1)*d)... [an = a1+(n-1)*d]
= 26/2 * [6 + 6 + (26-1)*8]
= 13*[12 + 25*8] = 13*[12+200] = 2756
Hence sum of 26 terms of given arithmetic sequence is 2756.
Hope it helps.
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