Question

How many terms of the series 9+12+15+...must be taken so that sum may be 306

Answer 1

Step-by-step explanation:

A/C to question,

9 + 12 + 15 + ....... n terms = 306

you can see that ,

common difference of the series is always equal to 3 . e.g (12 - 9) = (15 - 12) = 3

so, this series is in AP .

now,

Let total number of terms = n in such a way that sum of n terms = 306.

we know,

\begin{gathered}S_n=\frac{n}{2}[2a + (n-1)d]\\\\where, d\:is\: common \: difference\:a\:is\:1st\:term\end{gathered}

S

n

=

2

n

[2a+(n−1)d]

where,discommondifferenceais1stterm

here, a = 9 , d = 3 and Sn = 306

now,

\begin{gathered}306=\frac{n}{2}[2\times9+(n-1)3]\\\\306\times2=n[18+3n-3]\\\\612=15n+3n^2\\\\n^2+5n-204=0\\\\n=\frac{-5\pm\sqrt{25+816}}{2}\\\\n=12,-17\\\\hence \: \: \: \: \: \: n=12\end{gathered}

306=

2

n

[2×9+(n−1)3]

306×2=n[18+3n−3]

612=15n+3n

2

n

2

+5n−204=0

n=

2

−5±

25+816

n=12,−17

hencen=12

Answer 2

Answer :-

9 + 12 + 15 + ....... n terms = 306

you can see that ,

common difference of the series is always equal to 3 . e.g (12 - 9) = (15 - 12) = 3

so, this series is in AP .

so, this series is in AP .now,

Let total number of terms = n in such a way that sum of n terms = 306.

Let total number of terms = n in such a way that sum of n terms = 306.we know,

$s_{n} = \frac{n}{2}{2a + (n - 1)d}$

where d is common difference a is 1st term

here, a = 9 , d = 3 and Sn = 306

now,

$306 = \frac{n}{2}(2 \times 9 + (n - 1)3)$

306 × 2 = n [ 18 + 3n - 3 ]

$612 = 15n + {3n}^{2} \\ {n}^{2} + 5n - 204 = 0 \\ n = \frac{ - 5 + \sqrt{25 + 816} }{2}$

n = 12 , - 17

hence , n = 12

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