Question

Q. If the 6th and 10th
terms of an AP are 55 and
91 respectively Find
common difference. *​

Answer 1

Step-by-step explanation:

6th term 55

10th term 91

comman difference = 91-55÷10-6=36÷4=9

so comman difference =9

Answer 2

Given :

• 6th term of the AP = 55

• 10th term of the AP = 91

To Find :

The Common Difference of the AP.

Solution :

⠀⠀⠀⠀Let the common difference be d.

According to the laws of AP.

The terms of the AP can be found by using the formula :-

$\underline{\bf{t_{n} = a_{1} + (n - 1)d}}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀Where :

• $\sf{t_{n}}$ = nth term of the AP

• $\sf{a_{1}}$ = First term of the AP

• $\sf{d}$ = Common Difference

• $\sf{n}$ = No. of terms in the AP

Now , from the information , we can make two different Equation and by solving them , we can find the required value.

Equation (i) :

• nth term = 55
• n = 6

Using the formula for nth term and Substituting the values in it , we get :

$:\implies \bf{t_{n} = a_{1} + (n - 1)d}$

$:\implies \bf{55 = a_{1} + (6 - 1)d}$

$:\implies \bf{55 = a_{1} + 5d}$

$\therefore \bf{a_{1} + 5d = 55}\:\:\:\:[Equation.(i)]$

Hence, Equation (i) is $\bf{a_{1} + 5d = 55}$

Equation (ii) :

• nth term = 91
• n = 10

Using the formula for nth term and Substituting the values in it , we get :

$:\implies \bf{t_{n} = a_{1} + (n - 1)d}$

$:\implies \bf{91 = a_{1} + (10 - 1)d}$

$:\implies \bf{91 = a_{1} + 9d}$

$\therefore \bf{a_{1} + 9d = 91}\:\:\:\:[Equation.(ii)]$

Hence, Equation (ii) is $\bf{a_{1} + 4d = 55}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

Now , by subtracting the two Equations , we get :

$:\implies \bf{(a_{1} + 5d) - (a_{1} + 9d) = 55 - 91}$

$:\implies \bf{(a_{1} + 5d) - a_{1} - 9d = 55 - 91}$

$:\implies \bf{a_{1} + 5d - a_{1} - 9d = - 36}$

$:\implies \bf{\not{a_{1}} + 5d - \not{a_{1}} - 9d = - 36}$

$:\implies \bf{5d - 9d = - 36}$

$:\implies \bf{-4d = - 36}$

$:\implies \bf{\not{-}4d = \not{-} 36}$

$:\implies \bf{4d = 36}$

$:\implies \bf{d = \dfrac{36}{4}}$

$\therefore \bf{Common\:Difference\:(d) = 9}$

Hence, the common difference is 9.