Question

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.

Answer 1

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.