# Arithmetic Progression problems with solutions

We will discuss some arithmetic Progression problems with solutions in which students are facing problems while solving it.
1) Find the general term of the A.P. given by x + b, x + 3b, x + 5b,...
Solution : Here a = x + b ,
d = x + 3b - (x + b)
d = x + 3b - x - b
∴ d = 2b
The general term is given by the formula
$a_{n}$= a + (n - 1) d
$a_{n}$ = x + b + (n - 1)2b
∴ $a_{n}$ = x + (2n - 1)b

2) Show that $a^{2}, b^{2}, c^{2}$ are in A.P. if $\frac{1}{b + c}, \frac{1}{c + a}, \frac{1}{a + b}$ are in A.P.

Solution : Since $\frac{1}{b + c}, \frac{1}{c + a}, \frac{1}{a + b}$ are in A.P.

The common difference between the consecutive terms will be same
∴ $\frac{1}{c + a} - \frac{1}{b + c} = \frac{1}{a + b} - \frac{1}{c + a}$

$\frac{b - a}{(c + a)(b + c)} = \frac{c - b}{(c + a)(a + b)}$

∴ $\frac{b - a}{b + c} = \frac{c - b }{a + b}$

Cross multiply
(b - a)(b + a) = (c -b)(c + b)
i.e. $b^{2} - a^{2} = c^{2} - b^{2}$
⇒ $a^{2}, b^{2}, c^{2}$ are in A.P.

3) The 4th and 8th terms of an A.P. is 24 and the sum of the 6th and 10th terms is 34. Find the 1st term and the common difference of the A.P.
Solution : Here $a_{4} + a_{8}$ = 24 and $a_{6} + a_{10}$ = 34
a + 3d + a + 7d = 24 ------- [used the formula $a_{n}$ = a + (n - 1)d]
2a + 10d = 24
Divide by 2
a + 5d = 12 --------(1)
Now we have, $a_{6} + a_{10}$ = 34
a + 5d + a + 9d = 34
2a + 14d = 34
a + 7d = 17 --------(2)
Solving equation (1) and (2) simultaneously we get
d = $\frac{5}{2}$ and a = $\frac{-1}{2}$

4) The ratio of sums of 'm' and 'n' terms of an A.P is $m^{2} : n^{2}$. Show that the ratio of $m^{th}$ and $n^{th}$ term is 2m -1 : 2n - 1 .

Solution: Sum of n terms of an A.P. is given by
$S_{n} = \frac{n}{2}[2a + (n - 1)d]$

∴ $S_{m} = \frac{m}{2}[2a + (m - 1)d]$

As $\frac{S_{m}}{S_{n}} = \frac{m^{2}}{n^{2}}$

So, $\frac{m/2[2a + (m - 1)d]}{n/2[2a + (n - 1)d]} = \frac{m^{2}}{n^{2}}$

$\frac{2a + (m - 1)d}{2a + (n - 1)d} = \frac{m}{n}$

Cross multiply we get,
2an + n (m - 1)d = 2am + m(n - 1)d
2an - 2am = m(n - 1)d - n(m - 1)d
2a (n - m) = mnd - md - nmd + nd
2a (n - m) = - md + nd
2a (n - m) = d ( n - m)
∴ 2a = d
Now, $a_{m}$ = a + (m - 1)d
$a_{m}$ = a + (m - 1)2a
$a_{m}$ = a + 2am - 2a
∴ $a_{m}$ = 2am - a
Similarly $a_{n}$ = 2an - a
∴ the ratio of $a_{m}$ and $a_{n}$ will be
$\frac{a_{m}}{a_{n}} = \frac{2am - a}{2an - a}$

$\frac{a_{m}}{a_{n}} = \frac{2m - 1}{2n - 1}$ -----
proved

11th grade math

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