Question

The number of terms of the A.P $3,7,11,15,....$ to be taken so that the sum is $406$ is
$A)5$
$B)10$
$C)12$
$D)14$

Answer 1

Hint: While talking about the A.P and G.P, we need to know about the concept of Arithmetic and Geometric progression.
An arithmetic progression can be represented by $a,(a + d),(a + 2d),(a + 3d),...$where $a$ is the first term and $d$ is a common difference.
A geometric progression can be given by $a,ar,a{r^2},....$ where $a$ is the first term and $r$ is a common ratio.
Here in this given question, we clearly see that the difference is $4$ and they give arithmetic progress.
Formula used: The sum of the n-terms of the A.P is ${S_n} = (\dfrac{n}{2})(2a + (n - 1)d)$

Complete step-by-step solution:
Since from the problem given that we have, the number of terms of the A.P $3,7,11,15,....$
We will first find the first term and common difference of the A.P.
Since clearly, we see that the first term is $3$ which means $a = 3$ and the common difference can be found with the formula that the second term subtract the first term which means we have $d = b - a = 7 - 3 = 4$ is the common difference.
We need to find the sum of $406$ the occuring term. Hence, we use the sum of the n-terms formula where take ${S_n} = 406$ because which is the required term.
Thus, we have ${S_n} = (\dfrac{n}{2})(2a + (n - 1)d) \Rightarrow 406 = (\dfrac{n}{2})(2a + (n - 1)d)$ (where we need to find the $n$)
Since we have to $a = 3,d = 4$ substitute these values we get $406 = (\dfrac{n}{2})(2 \times 3 + (n - 1)4)$
Further solving we get $406 = (\dfrac{n}{2})(2 \times 3 + (n - 1)4) \Rightarrow 406 = (\dfrac{n}{2})(6 + 4n - 4)$
$406 = (\dfrac{n}{2})(2 + 4n) \Rightarrow 406 = n(1 + 2n)$
\[406 = n + 2{n^2} \Rightarrow 2{n^2} + n - 406 = 0\] which is like the quadratic equation, so we will use the quadratic formula to find its values.
$\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ where $b = 1,a = 2,c = - 406$
Hence, we get $\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} = \dfrac{{ - 1 \pm \sqrt {{1^2} - 4(2)( - 406)} }}{{2 \times 2}}$
Further solving we get \[\dfrac{{ - 1 \pm \sqrt {{1^2} - 4(2)( - 406)} }}{{2 \times 2}} = \dfrac{{ - 1 \pm \sqrt {{1^2} + 3248} }}{4}\]
\[\dfrac{{ - 1 \pm \sqrt {{1^2} + 3248} }}{4} = \dfrac{{ - 1 \pm \sqrt {3249} }}{4} \Rightarrow \dfrac{{ - 1 \pm 57}}{4} \Rightarrow 14,\dfrac{{ - 58}}{4}\]
Since we got two values but $\dfrac{{ - 58}}{4}$ is a negative value because it is the positive A.P of $3,7,11,15,....$ where $406$ cannot be negative.
Hence, we get $n = 14$, which means in fourteen terms the value $406$ will occur
Therefore, the option $D)14$ is correct.

Note: The quadratic equation is the two-degree equations which have at most two degrees, can be denoted as $a{x^2} + bx + c = 0$
In the quadratic equation, the value $a = 0$ is impossible, because if it happens then we get $bx + c = 0$ which is the linear function. So that it will never be zero.
Similarly, cubic equations have at most three degrees.