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# The fourth term of an arithmetic progression is $10$ and the eleventh term of it exceeds three times the fourth term by $1$. Find the sum of the first $20$ terms of the progression.

Last updated date: 27th Mar 2023
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Hint : Find first term and common difference to find the sum.
As we know that the nth term of an A.P. is denoted as ,
${a_n} = a + (n - 1)d$

From the question we can say that,
${a_4} = a + (4 - 1)d = a + 3d = 10\,\,\,\,...(i) \\ {a_{11}} = a + (11 - 1)d = a + 10d\,\,\,\, \\$
It is given that,
${a_{11}} = 3{a_4} + 1 \\ a + 10d = 3a + 9d + 1 \\ 2a - d + 1 = 0\,\,\,\,\,\,\,\,\,......(ii) \\$

On multiplying the equation $(ii)$ by 3 we get,
$6a - 3d + 3 = 0\,\,\,\,\,\,......(iii)$

Solving equation $(i)$ & $(iii)$we get,
$a = 1 \\ \& \\ d = 3 \\$
We know sum on n terms of an A.P. can be written as
${S_n} = \frac{n}{2}(2a + (n - 1)d) \\ {S_{20}} = \frac{{20}}{2}(2(1) + 19 \times 3) \\ {S_{20}} = 590 \\$
Hence, the sum of 20 terms of the series in 590.

Note :- In these types of questions of A.P. we have to first obtain an equation from the given data then solve the equation to get the unknowns like first term & common difference, after finding the unknowns, obtain the sum by using the formula of sum of an A.P.