How many terms of the serirs 9+12 +15 + .....must be taken to get the sum 306
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★ ARITHMETIC PROGRESSION ★
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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,
![S_n=\frac{n}{2}[2a + (n-1)d]\\\\where, d\:is\: common \: difference\:a\:is\:1st\:term](https://tex.z-dn.net/?f=S_n%3D%5Cfrac%7Bn%7D%7B2%7D%5B2a+%2B+%28n-1%29d%5D%5C%5C%5C%5Cwhere%2C+d%5C%3Ais%5C%3A+common+%5C%3A+difference%5C%3Aa%5C%3Ais%5C%3A1st%5C%3Aterm "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\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](https://tex.z-dn.net/?f=306%3D%5Cfrac%7Bn%7D%7B2%7D%5B2%5Ctimes9%2B%28n-1%293%5D%5C%5C%5C%5C306%5Ctimes2%3Dn%5B18%2B3n-3%5D%5C%5C%5C%5C612%3D15n%2B3n%5E2%5C%5C%5C%5Cn%5E2%2B5n-204%3D0%5C%5C%5C%5Cn%3D%5Cfrac%7B-5%5Cpm%5Csqrt%7B25%2B816%7D%7D%7B2%7D%5C%5C%5C%5Cn%3D12%2C-17%5C%5C%5C%5Chence+%5C%3A++%5C%3A++%5C%3A++%5C%3A++%5C%3A++%5C%3A+n%3D12 "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")
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,
here, a = 9 , d = 3 and Sn = 306
now,
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