Question

find first three term of an a.p whose 3rd term is 10 and 6th term is 19​

Answer 1

GIVEN :–

• 3rd term of A.P. = 10

• 6th term of A.P. = 19

TO FIND :–

• First three terms of A.P. = ?

SOLUTION :–

• We know that nth term of A.P. –

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

• Here –

$\\ \: \: \: \: \: { \huge{.}} \: \: { \bold{ a = first \:\: term }}$

$\\ \: \: \: \: \: { \huge{.}} \: \: { \bold{d = common \: \: difference }}$

• According to the first condition –

$\\ \implies { \bold{ T_{3} =10 }}$

$\\ \implies { \bold{ a + (3 - 1)d =10 }}$

$\\ \implies { \bold{ a + 2d =10 \: \: \: \: - - - eq.(1)}}$

• According to the second condition –

$\\ \implies { \bold{ T_{6} =19 }}$

$\\ \implies { \bold{ a + (6 - 1)d =19}}$

$\\ \implies { \bold{ a + 5d =19 \: }}$

• Using eq.(1) –

$\\ \implies { \bold{ (10 - 2d) + 5d =19 \: }}$

$\\ \implies { \bold{5d - 2d =19 - 10 \: }}$

$\\ \implies { \bold{3d =9 \: }}$

$\\ \implies { \bold{d = \cancel \dfrac{9}{3} \: }}$

$\\ \implies \large{ \boxed { \bold{d = 3}}}$

• Put the value of 'd' in eq.(1) –

$\\ \implies { \bold{ a + 2(3) =10}}$

$\\ \implies { \bold{ a =10 - 6}}$

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

$\\ \implies { \bold{first \: \:three \: \: terms = (a),(a + d),(a + 2d)}}$

$\\ \implies { \bold{first \: \:three \: \: terms = (4),(4 + 3),(4+ 2 \times 3)}}$

$\\ \implies { \bold{first \: \:three \: \: terms = 4,7,10}}$

Answer 2

Given:

• 3rd term of AP is 10.
• 6th term is 19

To Find:

• First three terms of AP.

Concept Used:

• We will use the general formula to find nth term if an AP .
• Then we will find first number and Common Difference.

Solution:

For answer refer to attachment.