Find the sum: 3+5+7+6+9+12+9+13+17.......30 terms

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Solution

The given sequence i.e., 3+5+7+6+9+12+9+13+17+……. to 30 terms.
Can be rewritten as 3+6+9+……. to 10 terms +7+12+17+…. to 10 terms
Clearly all these sequence forms an A.P. having 10 terms with first terms 3,5,7 and common different 3,4,5.
Hence, required sum = $\frac{10}{2}$
$2 \times 3 + (10 - 1) 3] + \frac{10}{2}$
$2 \times 5 + (10 - 1) 4] + \frac{10}{2}$
$[2 \times 7 + (10 - 1) 5]$
$= 5 (6 + 9 \times 3) + 5 (10 + 9 \times 4) + 5 (14 + 9 \times 5) = 5 (6 + 27 + 10 + 36 + 14 + 45)$
$= 5 \times 138 = 690$ .

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Similar questions

Find the sum of the series : 3 + 5 + 7 + 6 + 9 + 12 + 9 + 13 + 17 + .... to 3n terms.

Re-arrange suitably and find the sum in each of the following :

(i) $\frac{11}{12} + \frac{- 17}{3} + \frac{11}{2} + \frac{- 25}{2}$

(ii) $\frac{- 6}{7} + \frac{- 5}{6} + \frac{- 4}{9} + \frac{- 15}{7}$

(iii) $\frac{3}{5} + \frac{7}{3} + \frac{9}{5} + \frac{- 13}{15} + \frac{- 7}{3}$

(iv) $\frac{4}{13} + \frac{- 5}{8} + \frac{- 8}{13} + \frac{9}{13}$

(v) $\frac{2}{3} + \frac{- 4}{5} + \frac{1}{3} + \frac{2}{5}$

(vi) $\frac{1}{8} + \frac{5}{12} + \frac{2}{7} + \frac{7}{12} + \frac{9}{7} + \frac{- 5}{16}$

1 + 3 + 5 + 7 + 9

1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19

1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23

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