Question

the sum of first six terms of an AP is 42. The ratio of its 10th and its 30th term is 1:3. Calculate the first and the 13th term of the AP

Answer 1

EXPLANATION.

Sum of first 6 terms of an Ap = 42.

$\sf : \implies \: formula \: of \: sum \: of \: n \: terms \: of \: an \: ap \\ \\ \sf : \implies \: s_{n} \: = \frac{n}{2} (2a + (n - 1)d) \\ \\ \sf : \implies \: \: s_{6} \: = \frac{6}{2}(2a \: + 5d) = 42 \\ \\ \sf : \implies \: 3(2a + 5d) = 42\\ \\ \sf : \implies \: 6a \: + 15d \: = 42 \: \: \: \\ \\ \sf : \implies \: 2a \: + 5d \: = 14 \: \: \: .....(1)$

The ratio of it's 10th and 30th term is 1:3.

$\sf : \implies \: \dfrac{ t_{10} }{ t_{30}} = \dfrac{1}{3} \\ \\ \sf : \implies \: \frac{a + 9d}{a + 29d} = \frac{1}{3} \\ \\ \sf : \implies \: 3a \: + 27d \: = a \: + 29d \\ \\ \sf : \implies \: 2a \: = 2d \\ \\ \sf : \implies \: a \: = d \: \: \: .....(2)$

$\sf : \implies \: from \: equation \: (1) \: \: \: and \: \: \: (2) \: \: \: we \: get \\ \\ \sf : \implies \: 2a \: + \: 5a \: = 14 \\ \\ \sf : \implies \: 7a \: = 14 \\ \\ \sf : \implies \: a \: = 2 \\ \\ \sf : \implies \: d \: = 2$

=> 13th term of an Ap

=> a + 12d

=> 2 + 12 X 2

=> 26

Therefore,

First term = 2

13 th term = 26.

Answer 2

$\bf{\gray{\underbrace{\blue{GIVEN:-}}}}$

• The sum of first six terms of an AP is 42 .

• The ratio of its 10th and 30th term is 1:3 .

$\bf{\gray{\underbrace{\blue{TO\:FIND:-}}}}$

1. The first term of the A.P .
2. The 13th term of the AP .

$\bf{\gray{\underbrace{\blue{SOLUTION:-}}}}$

$\orange\bigstar\:\bf{\red{\overbrace{\underbrace{\purple{S_n\:=\:\dfrac{n}{2}\:[2a\:+\:(n\:-\:1)\:d]\:}}}}}$

$\green\bigstar\:\bf{\red{\overbrace{\underbrace{\purple{t_n\:=\:a\:+\:(n\:-\:1)\:d\:}}}}}$

Where,

• a = first term

• n = no. of term's .

• d = common difference .

✨ According to the question,

CASE - 1 :-

$\bf{:\implies\:S_6\:=\:\dfrac{6}{2}\:[2a\:+\:(6\:-\:1)\:d]\:}$

$\rm{:\implies\:42\:=\:3\:[2a\:+\:5d]\:}$

$\rm{:\implies\:2a\:+\:5d\:=\:\dfrac{42}{3}\:}$

$\bf\green{:\implies\:2a\:+\:5d\:=\:14\:}$----(1)

CASE - 2 :-

$\bf{:\implies\:\dfrac{t_{10}}{t_{30}}\:=\:\dfrac{1}{3}\:}$

$\rm{:\implies\:\dfrac{a\:+\:(10\:-\:1)\:d}{a\:+\:(30\:-\:1)\:d}\:=\:\dfrac{1}{3}\:}$

$\rm{:\implies\:\dfrac{a\:+\:9d}{a\:+\:29d}\:=\:\dfrac{1}{3}\:}$

$\rm{:\implies\:3a\:+\:27d\:=\:a\:+\:29d\:}$

$\rm{:\implies\:3a\:-\:a\:+\:27d\:-\:29d\:=\:0\:}$

$\rm{:\implies\:2a\:-\:2d\:=\:0\:}$

$\rm{:\implies\:2a\:=\:2d\:}$

$\bf\green{:\implies\:a\:=\:d\:}$

⍟ Now putting the value of a in the equation (1),

$\rm{:\implies\:2a\:+\:5d\:=\:14\:}$

$\rm{:\implies\:2d\:+\:5d\:=\:14\:}$

$\rm{:\implies\:7d\:=\:14\:}$

$\rm{:\implies\:d\:=\:\dfrac{14}{7}\:}$

$\bf\pink{:\implies\:d\:=\:2\:}$

$\bf\blue{:\implies\:a\:=\:2\:}$

⍟ Now, to calculate the 13th term of the A.P .

$\bf{:\implies\:t_{13}\:=\:a\:+\:(13\:-\:1)\:d\:}$

$\rm{:\implies\:t_{13}\:=\:2\:+\:12\times{2}\:}$

$\rm{:\implies\:t_{13}\:=\:2\:+\:24\:}$

$\bf\purple{:\implies\:t_{13}\:=\:26\:}$

___________________________

$\pink\therefore$ The first term of the A.P is "2" .

$\pink\therefore$ The 13th term of the A.P is "26" .