If 3-5-7- -nterms -5-8-11- - - - -10 terms -7- the value of n is-

The correct option is A 35 3 + 5 + 7 + .........+ n terms = n/2(6 + (n 1)2) = n/2 (2n + 4) 5 + 8 + 11 + ..........+10 terms = 5(10 + 9.3) = 185 n(n+2)/185 = ...

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Byju's Answer

Standard XI

Mathematics

Sigma n3

If 3+5+7+…… ....

Question

If $\frac{3 + 5 + 7 + . . . . . . n t e r m s}{5 + 8 + 11 + . . . . . . . . . + 10 t e r m s}$ = 7, the value of n is:

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Solution

The correct option is A
35

3 + 5 + 7 + .........+ n terms = $\frac{n}{2}$ (6 + (n-1)2) = $\frac{n}{2}$ (2n + 4)

5 + 8 + 11 + ..........+10 terms = 5(10 + 9.3) = 185

$\frac{n (n + 2)}{185}$ = 7 ⇒ n = 35

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If $\frac{3 + 5 + 7 + . . . . . . n t e r m s}{5 + 8 + 11 + . . . . . . . . . + 10 t e r m s}$ = 7, the value of n is:

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If 3+5+7+......nterms5+8+11+.........+10terms = 7, the value of n is:

The correct option is A 35 3 + 5 + 7 + .........+ n terms = n2(6 + (n-1)2) = n2 (2n + 4) 5 + 8 + 11 + ..........+10 terms = 5(10 + 9.3) = 185 n(n+2)185 = 7 ⇒ n = 35

If $\frac{3 + 5 + 7 + . . . . . . n t e r m s}{5 + 8 + 11 + . . . . . . . . . + 10 t e r m s}$ = 7, the value of n is:

The correct option is A
35

3 + 5 + 7 + .........+ n terms = $\frac{n}{2}$ (6 + (n-1)2) = $\frac{n}{2}$ (2n + 4)

5 + 8 + 11 + ..........+10 terms = 5(10 + 9.3) = 185

$\frac{n (n + 2)}{185}$ = 7 ⇒ n = 35