Question

if S6 = 42 the ratio of is 10 term to 30 term is 1:3 . Find the 1 and 13 terms of an AP​

Answer 1

Answer:

First term is 7 and 13th term is 91.

Step-by-step explanation:

From the properties of AP :

• nth term = a + ( n - 1 )d , where a is the first term, n is the number of terms and d is the common difference between the terms.
• S${}_6$ = ( n / 2 ){ 2a + ( n - 1 )d }

Let : first term of this AP be a and common difference between the terms be d.

According to the given conditions :

= > Sum of first 6 terms = 42

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

= > 3( 2a + 5d ) = 42

= > 2a + 5d = 42 / 3 = 14 ...( 1 )

Also : Ratio of 10th term to 30th term us 1 : 3

= > { a + ( 10 - 1 )d } : { a + ( 30 - 1 )d } = 1 : 3

= > ( a + 9d ) : ( a + 29d ) = 1 : 3

= > 3( a + 9d ) = a + 29d

= > 3a + 27d = a + 29d

= > 3a - a = 29d - 27d

= > 2a = 2d

= > a = d

Thus, in ( 1 )

= > 2a + 5d = 14

= > 2a + 5a = 14

= > 7a = 14

= > a = 7

Therefore : 13th term = a + ( 13 - 1 )d = 7 + 12( 7 ) = 13( 7 ) = 91.

Answer 2

$\Large{\textbf{\underline{\underline{According\:to\:the\:Question}}}}$

Assumption

First term be a

Also

Common difference be d

Hence

Sum of 6 terms = 42

Using Formula :-

$\tt{\rightarrow\dfrac{6}{2}[2a+(6-1)d]=42}$

3(2a + 5d) = 42

Therefore,

$\tt{\rightarrow 2a+5d=\dfrac{42}{3}}$

(2a + 5d) = 14 ..... (1)

Therefore,

${\boxed{\sf\:{Ratio\;of\;10t\;term\;to\;30th\; term}}}$

= 1 : 3

[a + (10 - 1)d] : [a + (30 - 1)d] = 1 : 3

(a + 9d) : (a + 29d) = 1 : 3

3(a + 9d) = a + 29d

3a + 27d = a + 29d

3a - a = 29d - 27d

2a = 2d

Hence,

a = d

From (1) we have :-

2a + 5d = 14

2a + 5a = 14

7a = 14

$\tt{\rightarrow a=\dfrac{14}{7}}$

a = 7

Hence,

13th term = a + (13 - 1)d

= 7 + 12(7)

= 13 × 7

= 91