Let a_n be the n^th term of an A.P.

Question

Let $a_n$ be the $n^{th}$ term of an A.P. If $S_n = a_1 + a_2 + a_3 + \ldots + a_n = 700$, $a_6 = 7$ and $S_7 = 7$, then $a_n$ is equal to :

Accepted Answer

Sum of n terms of an AP ($S_n$):  $$S_n = \frac{n}{2} [2a + (n-1)d]$$ where 'a' is the first term and 'd' is the common difference.

nth term of an AP ($a_n$):  $a_n = a + (n-1)d$

Given Information:

$S_n = 700$ (The sum of the first n terms is 700)

$a_6 = 7$ (The 6th term is 7)

$S_7 = 7$ (The sum of the first 7 terms is 7)

Steps to Solve:

Use $S_7$ to find a relationship between 'a' and 'd':

$$S_7 = \frac{7}{2} [2a + (7-1)d] = 7$$

$$\frac{7}{2} [2a + 6d] = 7$$

$2a + 6d = 2$

$a + 3d = 1$  (Equation 1)

Use $a_6$ to find another relationship between 'a' and 'd':

$a_6 = a + (6-1)d = 7$

$a + 5d = 7$  (Equation 2)

Solve the system of equations (Equation 1 and Equation 2) to find 'a' and 'd':

Subtract Equation 1 from Equation 2:

$(a + 5d) - (a + 3d) = 7 - 1$

$2d = 6$

$d = 3$

Substitute d = 3 into Equation 1:

$a + 3(3) = 1$

$a + 9 = 1$

$a = -8$

Find 'n' using $S_n = 700$:

$$S_n = \frac{n}{2} [2a + (n-1)d] = 700$$

$$\frac{n}{2} [2(-8) + (n-1)(3)] = 700$$

$$n[-16 + 3n - 3] = 1400$$

$$n[3n - 19] = 1400$$

$3n^2 - 19n - 1400 = 0$

Solve the quadratic equation for 'n':

We can use the quadratic formula: $$n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

$$n = \frac{19 \pm \sqrt{(-19)^2 - 4(3)(-1400)}}{2(3)}$$

$$n = \frac{19 \pm \sqrt{361 + 16800}}{6}$$

$$n = \frac{19 \pm \sqrt{17161}}{6}$$

$$n = \frac{19 \pm 131}{6}$$

We have two possible values for n:

$$n = \frac{19 + 131}{6} = \frac{150}{6} = 25$$

$$n = \frac{19 - 131}{6} = \frac{-112}{6} = -\frac{56}{3}$$ (Since 'n' must be a positive integer, we discard this solution)

So, $n = 25$

Find $a_n$ (which is $a_{25}$):

$a_{25} = a + (25-1)d$

$a_{25} = -8 + (24)(3)$

$a_{25} = -8 + 72$

$a_{25} = 64$

Therefore, $a_n = 64$.  The correct answer is Option D.

Solutions · 1