Question

©30 points© . For maths lover .


The sum of first 6 terms of an AP is 42 . The ratio of its 10th term to 30th term is 1 : 3 . Find the first and the 13th term of the AP .

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Answer 1

Answer :-

→ The first term is 2 and 13th term term is 26 .

Step-by-step explanation :-

$\huge \pink{ \mid{ \underline{ \overline{ \tt Solution :- }} \mid}}$

→ Let a be the first term and d be the common difference of the given AP . Then,

→ $\sf a_{10} = a + 9d \: \: and \: \: a_{30} = a + 29d$

$\begin{lgathered}\sf \therefore \frac{a_{10}}{a_{30}} = \frac{1}{3} . \\ \\ \sf \implies \frac{a + 9d}{a + 29d} = \frac{1}{3} . \\ \\ \sf \implies3a + 27d = a + 29d. \\ \\ \sf \implies \cancel2a = \cancel2d. \\ \\ \sf \large \implies a = d. \\ \\ \\ \sf Also, S_n = \frac{n}{2} \bigg(2a + (n - 1)d \bigg). \\ \\ \sf \implies S_6 = \frac{6}{2} (2a + 5d). \\ \\ \sf = 3(2a + 5d). \\ \\ \sf = (6a + 15d). \\ \\ \sf = (6a + 15a). \: \: \: \: \{ \because d = a \} \\ \\ \large \sf = 21a.\end{lgathered}$

$\begin{lgathered}\sf But, S_6 = 42 . ( given ) .\\ \\ \sf \therefore 21a = 42 . \\ \\ \huge \orange{ \boxed{ \sf \implies a = 2.}}\end{lgathered}$

Thus, a = 2 and d = 2 .

∴ 13th term , [tex] a_13 [/tex]

= ( a + 12d ) .

= ( 2 + 12 × 2 ) .

$\huge \blue { \boxed{ \sf = 26 . }}$

Hence, the first term is 2 and 13th term term is 26 .

Answer 2

Here is the solution of your problem buddy ↙️↙️↙️↙️☺️☺️✌️✌️

S6 =42

a + 9d                    1

------------     =      

a + 29d                  3

cross multiply we  get

3a + 27d = a +29 d

2a - 2d = 0 -------(1)

 

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)

solve eq 1 and 2

2a - 2d = 0

2a +5d = 14

we get 

d= 2

a = 2

13th term of AP

= a + (n-1)d

2+ (13-1) 2

= 2+ 24

=26

Hope this helps you buddy ☺️☺️❤️❤️✌️✌️

$<marquee behaviour=slide bg color= pink > <h1> All the best ☺️☺️♥️♥️</h1> </marquee>$