Question

9th term of an arithmetic progression is 19 the sum of 4th and 7th term is 24 find Arithmetic progression​

Answer 1

Given :-

9th term ($T_{9}$) of the A.P. = 19

Sum of 4th ($T_{4}$) and 7th term ($T_{7}$) of the A.P. = 24

Solution :-

$T_{n}$ = 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.

⇒ $T_{9}$ = a + (9-1) d

⇒ 19 = a + 8d

⇒ 19 - 8d = a       ....i.)

⇒ $T_{4}$ = a + (4-1) d

⇒ $T_{4}$ = a + 3d

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

⇒ $T_{4}$ = 19 - 8d + 3d

⇒ $T_{4}$ = 19 - 5d

⇒ $T_{7}$ = a + (7-1) d

⇒ $T_{7}$ = a + 6d

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

⇒ $T_{7}$ = 19 - 8d  + 6d

⇒ $T_{7}$ = 19 - 2d

Sum of 4th ($T_{4}$) and 7th term ($T_{7}$) = 24

⇒ $T_{4}$ + $T_{7}$ = 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 = $\frac{-14}{-7}$

⇒ 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.

Answer 2

Answer:

in above answer where's come that a=19-8d ?

Step-by-step explanation:

answer fast plz anyone