Question

The ${17}^{\text{th}}$ term of an A·P exceeds its ${10}^{\text{th}}$ term by 7. Find the common difference.

Answer 1

Hint:
The ${n}^{\text{th}}$ term of an AP is given by ${a_n} = a + (n – 1) d$.
where a = first term of an A.P and d is a common difference.
Using the ${n}^{\text{th}}$ formula, write ${17}^{\text{th}}$ and ${10}^{\text{th}}$ term and form an expression according to question.

Complete step by step solution:
We know that the ${n}^{\text{th}}$ term of an A·P is given by, ${a_n} = a + (n – 1) d$,
where ‘a’ is the first term of an A·P and ‘d’ is a common difference.
A·P can be represented by:-
a, a + d, a + 2d, a + 3d, . . .
According to the question;
${17}^{\text{th}}$ term of A·P exceeds its ${10}^{\text{th}}$ term by 7.
So, ${17}^{\text{th}}$ term of A·P $= a + (17 – 1)d$
$\Rightarrow {a_{17}} = a + 16d$ ------ (1)
and the ${10}^{\text{th}}$ term of A·P $= a + (10 – 1)d$
$\Rightarrow {a_{10}} = a + 9d$ ------ (2)
According to question;
${17}^{\text{th}}$ term of an A·P exceeds its ${10}^{\text{th}}$ term by 7.
${a_{17}} – {a_{10}} = 7$
$\Rightarrow (a + 16d) – (a + 9d) = 7$
$\Rightarrow a + 16d – a – 9d = 7$
$\Rightarrow 16d – 9d = 7$
$\Rightarrow 7d = 7$
$\Rightarrow d = \dfrac{7}{7}$
$\Rightarrow d = 1$

∴ common difference of the given A·P is 1.

Note:
The general form of an Arithmetic Progression is a, a + d, a + 2d, a + 3d and so on. Thus, ${n}^{\text{th}}$ term of an AP series is ${{\rm{T}}_{\rm{n}}} = a + (n - 1) d$, where ${T_n} = {n}^{\text{th}}$ term and a = first term. Here d = common difference = ${{\rm{T}}_{\rm{n}}} - ({{\rm{T}}_{\rm{n-1}}} )$.