Question

the sum of the first 6 terms of an AP is 42.The ratio of the 10th term is to the 13th term is 1:3,.Find the 1st and the 13th term of an AP​

Answer 1

$\huge\bold\pink{SOLUTION:}$

S6 =42

$\frac{a + 9d}{a + 2ad} = \frac{1}{3}$

cross multiply we get

3a + 27d = a +29 d

2a - 2d = 0 __________________ (1)

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its given that

sum of first six terms of an AP is 42

therefore

S6 = n/2 ( 2a + (n-1) d)

42 = 6/2 ( 2a + (6-1) d)

42 = 3 (2a + 5d )

14 = 2a +5d

2a +5d = 14 ________ (2)

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solve eq 1 and 2

2a - 2d = 0

2a +5d = 14

we get

d= 2

a = 2

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13th term of AP

= a + (n-1)d

2+ (13-1) 2

= 2+ 24

=26

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Answer 2

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