Question

9 term of an Arithmetic Progression is 19.
The sum of 4th and 7th term is 24. Find the
Arithmetic Progression.​

Answer 1

$\underbrace{\underline{\bf{\bigstar\:Required\:Answer:-}}}$

➡ Given :

• 9th Term of the A.P. = 19
• 4th Term + 7th Term = 24

➡ To Find :

• The Arithmetic Progression (AP).

➡ Solution :

We know that the $\sf n^{th}$ term of an AP is generalised as -

$\leadsto \: \boxed{ \sf a _{n} =a + (n - 1)d }$

Here,

• a = first term of the AP
• d = common difference
• n = no. of terms
• an = n th term

$\textit{\textbf{\dag\:According\:to\:the\:Question:}}$

$\colon \longrightarrow \sf \: 9 {}^{th} \: term = 19$

$\colon \longrightarrow \sf \: a + 8d = 19$

$\colon \longrightarrow \: \boxed{ \sf{a = 19 - 8d}} \sf \: ...(i)$

And..

$\colon \longrightarrow \sf \: 4 {}^{th} \: term + {7}^{th} \: term = 24$

$\colon \longrightarrow \sf \: a + 3d + a + 6d = 24$

$\colon \longrightarrow \sf \: 2a + 9d = 24$

• From eq(i) , a = 19 - 8d

$\colon \longrightarrow \sf \: 2(19 - 8d) + 9d = 24$

$\colon \longrightarrow \sf \: 38 - 16d + 9d = 24$

$\colon \longrightarrow \sf \: - 7d = 24 - 38$

$\colon \longrightarrow \sf \: \cancel{ - 7d }= \cancel{ - 14}$

$\colon \implies \underline{ \boxed{ \bf{ \red{d = 2}}}} \: \bigstar$

• Put it in eq(i)

$\\ \colon \longrightarrow \sf \: a = 19 - 8 \times 2$

$\colon \implies \: \underline{ \boxed{ \bf{ \purple{a = 3}}}} \: \bigstar$

$\underline{\dag\:\bf Arithmetic \:Progression :-}$

» Arithmetic Progression is in the form :

✒ a , a + d , a + 2d , a + 3d ,...

» For this question,

• a = 3
• d = 2

So,

Required Arithmetic Progression would be :

$\sf{ 3,3+2, 3+2\times2, 3+3\times2,...}$

$\therefore\underline{\sf A. P. = \pink{3, 5,7,9,...}}$

_____________________________

Answer 2

Answer:

I hope it is helpful

Step-by-step explanation:

Given :-

9th term (T_{9}T

9

) of the A.P. = 19

Sum of 4th (T_{4}T

4

) and 7th term (T_{7}T

7

) of the A.P. = 24

Solution :-

T_{n}T

n

= a + (n-1)d

where,

a = the first term of the A.P.

n = number of terms of the A.P.

d = difference of consecutive terms of the A.P.

⇒ T_{9}T

9

= a + (9-1) d

⇒ 19 = a + 8d

⇒ 19 - 8d = a ....i.)

⇒ T_{4}T

4

= a + (4-1) d

⇒ T_{4}T

4

= a + 3d

[Substituting the value of a from eq. i.)]

⇒ T_{4}T

4

= 19 - 8d + 3d

⇒ T_{4}T

4

= 19 - 5d

⇒ T_{7}T

7

= a + (7-1) d

⇒ T_{7}T

7

= a + 6d

[Substituting the value of a from eq. i.)]

⇒ T_{7}T

7

= 19 - 8d + 6d

⇒ T_{7}T

7

= 19 - 2d

Sum of 4th (T_{4}T

4

) and 7th term (T_{7}T

7

) = 24

⇒ T_{4}T

4

+ T_{7}T

7

= 24

[Substituting the value of the terms obtained above]

⇒ (19 - 5d) + (19 - 2d) = 24

⇒ 19 - 5d + 19 - 2d = 24

⇒ 38 - 7d = 24

⇒ - 7d = 24 -38

⇒ - 7d = -14

⇒ d = \frac{-14}{-7}

−7

−14

⇒ d = 2

⇒ a = 19 - 8d

⇒ a = 19 - 8 x 2

⇒ a = 19 - 16

⇒ a = 3

First term = a = 3

Second term = a + d = 3 + 2 = 5

Third term = a + 2d = 3 + 2 x 2 = 3 + 4 = 7

Fourth term = a + 3d = 3 + 3 x 2 = 3 + 6 = 9

Fifth term = a + 4d = 3 + 4 x 2 = 3 + 8 = 11

Sixth term = a + 5d = 3 + 5 x 2 = 3 + 10 = 13

Seventh term = a + 6d = 3 + 6 x 2 = 3 + 12 = 15

The A.P. is -

3, 5, 7, 9, 11, 13, 15.