Question

If 9th term & 21st term of an A.P. are 75 & 183 respectively, then find 41st term of that A.P.

plz ans......​

Answer 1

Given :

• 9th term of the AP = 75.

• 21st term of the AP = 183.

To find :

41st term of the AP.

Solution :

Let the First term of the AP be a and the common difference be d.

First let us find the nth term of the 9th term and 21st term.

We know the formula for nth term of an AP i.e,

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

Where :

• tn = nth term of the AP.

• d = Commom Difference.

• n = No. of terms.

• a = First term.

Now using the above equation and substituting the values in it, we get :

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

$:\implies \bf{75 = a_{1} + (9 - 1)d}$

$:\implies \bf{75 = a_{1} + 8d}$

$\boxed{\therefore \bf{a_{1} + 8d = 75}}$⠀⠀⠀⠀⠀⠀⠀⠀⠀Eq.(i)

We know the formula for nth term of an AP i.e,

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

Where :

• tn = nth term of the AP.

• d = Commom Difference.

• n = No. of terms.

• a = First term.

Now using the above equation and substituting the values in it, we get :

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

$:\implies \bf{183 = a_{1} + (20 - 1)d}$

$:\implies \bf{183 = a_{1} + 20d}$

$\boxed{\therefore \bf{a_{1} + 20d = 183}}$⠀⠀⠀⠀⠀⠀⠀⠀⠀Eq.(ii)

By subtracting Eq.(ii) from Eq.(i) , we get :

$:\implies \bf{(a_{1} + 8d) - (a_{1} + 20d) = 75 - 183}$

$:\implies \bf{(a_{1} + 8d) - (a_{1} + 20d) = - 108}$

$:\implies \bf{a_{1} + 8d - a_{1} - 20d = - 108}$

$:\implies \bf{\not{a_{1}} + 8d - \not{a_{1}} - 20d = - 108}$

$:\implies \bf{8d - 20d = - 108}$

$:\implies \bf{- 12d = - 108}$

$:\implies \bf{\not{-} 12 = \not{-} 108}$

$:\implies \bf{12 = 108}$

$:\implies \bf{d = \dfrac{108}{12}}$

$:\implies \bf{d = 9}$

$\boxed{\therefore \bf{d = 9}}$

Hence the common difference of the AP is 9.

Now Substituting the value of common difference (d) in the Eq.(i) , we get :

$:\implies \bf{a_{1} + 8d = 75}$

$:\implies \bf{a_{1} + 8(9) = 75}$

$:\implies \bf{a_{1} + 72 = 75}$

$:\implies \bf{a_{1} = - 72 + 75}$

$:\implies \bf{a_{1} = 3}$

$\boxed{\therefore \bf{a_{1} = 3}}$

Hence the First term of the AP is 3.

Now to find the 41st term of the AP :-

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

Where :

• tn = nth term of the AP.

• d = Commom Difference.

• n = No. of terms.

• a = First term.

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

$:\implies \bf{t_{41} = 3 + (41 - 1)9}$

$:\implies \bf{t_{41} = 3 + (40)9}$

$:\implies \bf{t_{41} = 3 + 360}$

$:\implies \bf{t_{41} = 363}$

$\boxed{\therefore \bf{t_{41} = 363}}$

Hence the 41st term of the AP is 363.

Answer 2

Answer:

Hope it will help you...