Question

1. There are four term in A.P. The sum of first three terms is 42 & the last 3 terms is 54. Find terms of ap​

Answer 1

Answer:

Four terms are: a-3d,a-d,a+d,a+3d

First three terms is 42

a-3d+a-d+a+d=42

3a-3d=42

a-d=14...........(A)

Last three terms is 54

a-d+a+d+a+3d=54

3a+3d=54

a+d=18.......... (B)

from (A)&(B)

2a=32

a=16

put the value of a in equation A

a-d=14

16-14=d

2=d

a=16,d=2

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Answer 2

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AP : 10, 14, 18, 22

Given :

• Number of terms in an AP = 4
• Sum of first three terms of AP = 42
• Sum of last three terms of AP = 54

To Find :

• Terms in the AP.

Solution :

a$_n$ = a + (n - 1) d

a$_1$ = a

a$_2$ = a + d

a$_3$ = a + 2d

a$_4$ = a + 3d

Given that,

a$_1$ + a$_2$ + a$_3$ = 42

$\implies$a + (a + d) + (a + 2d) = 42

$\implies$ 3a + 3d = 42

$\implies$ a + d = 14 ------ equation 1

a$_2$ + a$_3$ + a$_4$ = 54

$\implies$(a + d) + (a + 2d) + (a + 3d) = 54

$\implies$3a + 6d = 54

$\implies$ 3(a + 2d) = 54

$\implies$a + 2d = 18 -------- equation 2

subtract equation 1 from 2

(a + 2d) - (a + d) = 18 - 14

a + 2d - a - d = 4

d = 4

Substitute d = 4 in equation 1

a + d = 14

a + 4 = 14

a = 10

Therefore, first term of AP = 10

second term = 10 + 4 = 14

third term = 14 + 4 = 18

forth term = 18 + 4 = 22

Therefore, AP : 10, 14, 18, 22