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Find x, if 1+6+11+16+...+x=148.

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Find $x$ , if $1 + 6 + 11 + 16 + ... + x = 148$ .

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Find

x

, if

1 + 6 + 11 + 16 + ... + x = 148

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Text solutionVerified

Here the series is an

A . P

with

a = 1, d = 6 - 1 = 11 - 6 = 16 - 11 = 5

1 + 6 + 11 + 16 + ... + x = 148

Let there be n terms in the series.Them,

\frac{n}{2} [2 a + (n - 1) d] = 148

\frac{n}{2} [2 + (n - 1) \times 5] = 148

5 n^{2} - 3 n - 296 = 0

(n - 8) (5 n + 37) = 0

n = 8

since

n \neq < 0

x = n^{t h}

term

x = a + (n - 1) d = 1 + (8 - 1) \times 5 = 36

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