Question

the first term of two aps are equal and the ratios of their common difference is 1:2 .of the 7th term of the first ap is 28 and the 8th term of the 2nd APis 29.find the two AP's​

Answer 1

GIVEN :–

• First term(a) of two A.P.'s are equal.

• Ratio of their common difference(d) is 1:2.

• 7th term of the First A.P. is 28.

• 8th term of the Second A.P. is 29 .

TO FIND :–

• Both A.P.'s = ?

SOLUTION :–

• Let first term of both A.P's are 'a' , common difference of first A.P. is 'd' and Common difference of Second A.P. is '2d' .

• We know that nth term of A.P. is –

$\\ \longrightarrow \large { \boxed{ \bold{ T_{n} = a + (n - 1)d }}}$

• 7th term of the First A.P. = 28.

$\\ \implies { \bold{ T_{7} = 28 }}$

$\\ \implies { \bold{ a + (7 - 1)d = 28 }}$

$\\ \implies { \bold{ a + 6d = 28 ----eq.(1) }}$

• 8th term of the Second A.P. = 29

$\\ \implies { \bold{ T_{8} = 29 }}$

$\\ \implies { \bold{a + (8 - 1)(2d)= 29 }}$

$\\ \implies { \bold{a + 14d= 29 \: \: \: - - - - eq.(2) }}$

• Put the value of 'a' from eq.(1) in eq.(2) –

$\\ \implies { \bold{(28 - 6d) + 14d= 29 }}$

$\\ \implies { \bold{28 + 8d= 29 }}$

$\\ \implies { \bold{8d= 1 }}$

$\\ \implies \large { \boxed{ \bold{d= \dfrac{1}{8} }}}$

• Now using eq.(1) –

$\\ \implies { \bold{ a + 6( \dfrac{1}{8}) = 28 }}$

$\\ \implies { \bold{ a + \dfrac{3}{4} = 28 }}$

$\\ \implies { \bold{ a = 28 - \dfrac{3}{4} }}$

$\\ \implies \large{ \boxed{ \bold{ a = \dfrac{109}{4} }}}$

▪︎ So that –

• First A.P. –

$\implies{ \bold{ \dfrac{109}{4} , \dfrac{219}{8} , \dfrac{55}{2} ...... \: }}$

• Second A.P. –

$\implies{ \bold{ \dfrac{109}{4} , \dfrac{55}{2} , \dfrac{111}{4} ...... \: }}$