Question

The sum of the 4th and 8th term of an ap is 24 and the sum of 6th and 10th term of an ap is 34 and find the first three terms of the AP and the 29th term of an ap

Answer 1

GIVEN:

• Sum of 4th and 8th term = 24
• Sum of 6th and 10th term = 34

TO FIND:

• First term ( a ) , second term ( t2 ) , third term ( t3 ) & 29th term ( t29 )

SOLUTION:

Given,

t4 + t8 = 24

t6 + t10 = 34

We know that

tn = a + ( n - 1 )d

Where

• tn : nth term
• a : first term
• n : no. of terms
• d : common difference

a + (4 - 1)d + a + (8 - 1)d = 24

2a + 3d + 7d = 24

2a + 10d = 24

2(a + 5d) = 24

a + 5d = 12

t6 = 12

Assuming as eq 1

_______________________

a + (6 - 1)d + a + (10 - 1)d = 34

2a + 14d = 34

2(a + 7d) = 34

a + 7d =17

t8 = 17

Assuming as eq 2

Now, eq 2 - 1

a + 7d = 17

a + 5d = 12

(-) (-)

2d = 5

Common difference = 5/2

Substituting d = 5/2 in eq 1

a + 5(5/2) = 12

a + 25/2 = 12

a = 12 - 25/2

a = -1/2

__________________________

Now,

t2 = a + (2 - 1)d

t2 = -1/2 + 5/2

t2 = 5 - 1/2 = 4/2

t2 = 2

_____________

t3 = a + (3 - 1)d

t3 = -1/2 + 2(5/2)

t3 = - 1 + 10/2

t3 = 9/2

_____________

t29 = a + (29 - 1)d

t29 = -1/2 + 28(5/2)

t29 = -1/2 + 70

t29 = 139/2

ANSWERS

★ a = -1/2

★ t2 = 2

★ t3 = 9/2

★ t29 = 139/2

Answer 2

GIVEN :–

• Sum of the 4th and 8th term of an A.P. is 24.

• Sum of 6th and 10th term of an A.P. is 34.

TO FIND :–

• First three terms = ?

• 29th term of an A.P. = ?

SOLUTION :–

• If first term of A.P. is a , Common difference is d and total term of A.P. is n then nth term –

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

• According to the first condition –

$\\ \implies \: { \sf{ T_{4} + T_{8} = 24}}$

$\\ \implies \: { \sf{ a + (4 - 1)d + a + (8 - 1)d = 24}}$

$\\ \implies \: { \sf{ a + 3d + a + 7d = 24}}$

$\\ \implies \: { \sf{ 2a + 10d = 24}}$

$\\ \implies \: { \sf{ 2(a + 5d) = 24}}$

$\\ \implies \: { \sf{ a + 5d = 12 \: \: \: \: \: \: \: \: \: \: - - - - eq.(1)}}$

• According to the second condition –

$\\ \implies \: { \sf{ T_{6} + T_{10} = 34}}$

$\\ \implies \: { \sf{ a + (6 - 1)d + a + (10 - 1)d = 34}}$

$\\ \implies \: { \sf{ a + 5d + a + 9d = 34}}$

$\\ \implies \: { \sf{ 2a + 14d = 34}}$

$\\ \implies \: { \sf{ 2(a + 7d) = 34}}$

$\\ \implies \: { \sf{ a + 7d = 17}}$

• Now using eq.(1) –

$\\ \implies \: { \sf{ (12 - 5d) + 7d = 17}}$

$\\ \implies \: { \sf{ 12 + 2d= 17}}$

$\\ \implies \: { \sf{ 2d= 17 - 12}}$

$\\ \implies \large { \boxed{ \sf{ d= \dfrac{5}{2} }}}$

• Now put the value of 'd' in eq.(1) –

$\\ \implies { \sf{ a + 5 \left( \frac{5}{2} \right )= 12 }}$

$\\ \implies { \sf{ a + \left( \frac{25}{2} \right )= 12 }}$

$\\ \implies { \sf{ a = 12 - \left( \frac{25}{2} \right ) }}$

$\\ \implies \large { \boxed{ \sf{ a= - \dfrac{1}{2} }}}$

$\\ {\huge{.}} \: \: { \sf{ First \: \: three \: \: terms \: \: are \implies - \dfrac{ 1}{2} \: , \: 2 \: , \: \dfrac{9}{2} }}$

• Now Let's find 29th term –

$\\ \implies \: { \sf{ T_{29} = - \dfrac{ 1}{2} + (29 - 1) \dfrac{5}{2} }}$

$\\ \implies \: { \sf{ T_{29} = - \dfrac{ 1}{2} + ( \cancel{28}) \dfrac{5}{ \cancel{2}} }}$

$\\ \implies \: { \sf{ T_{29} = - \dfrac{ 1}{2} + 14 \times 5}}$

$\\ \implies \: { \sf{ T_{29} = - \dfrac{ 1}{2} + 70}}$

$\\ \longrightarrow \: \large { \boxed{ \sf{ T_{29} = \dfrac{ 139}{2}} }}$