Question

in an A.P S4=42 and S3=24. then find the fourth term fo an A.P
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Answer 1

Step-by-step explanation:

given

$s_{4} = 42 \: and \: s _{3} = 24$

we know that

$s_{n} = \frac{n}{2} (2a + (n - 1)d)$

so

$s_{4} = \frac{4}{2} (2a + (4 - 1)d \\ 42 = 2(2a + 3d) \: \: \: \: \: \: \: \: \: \: \\ \frac{42}{2} = (2a + 3d) \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ {21} = 2a +3 d - - - ( \: i \: )$

also,

$s_{3} = \frac{3}{2} (2a + (3- 1)d) \\ 24 = \frac{3}{2} (2a + 2d ) \\ \\ 24 = \frac{3}{2} 2(a + d ) \\ \frac{24}{3} = a + d \\ 8 = a + d \: \: \\ a = 8 - d - - - ( \: ii \: )$

take ( i )

21=2a + 3d

21=2(8-d)+3d. [from ii]

21= 16+3d-2d

d=5

take eq ( ii )

a=8-5

a=3

$a_{4} = a + (n - 1) d \\ = 3 +( 4 - 1)5 \\ = 3 + 3 \times 5 \\ = 3 + 15 \\ = 18$

therefore fourth term of A.P IS 18