Question

the sum of the first six terms of an arithmetic progression is 42.The ratio of its 10th term to its 30th term is 1:3.calculate the first and the thirteen term of the AP

Answer 1

hope you will be satisfied with my answer..

Answer 2

☯ Let a be the first term and d be the common difference of the given AP.

We know that,

$\star\;{\boxed{\sf{\pink{S_n = \dfrac{n}{2} \bigg( 2a + (n - 1)d \bigg)}}}}$

Therefore,

$:\implies\sf S_6 = \dfrac{6}{2} \bigg(2a + (6 - 1)d \bigg)$

$:\implies\sf 42 = \cancel{ \dfrac{6}{2}} \bigg(2a + 5d \bigg)$

$:\implies\sf 42 = 3(2a + 5d)$

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

$:\implies\sf 2a + 5d = 14\qquad\qquad\bigg\lgroup\bf eq\;(1) \bigg\rgroup$

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☯ The ratio of its 10th term to its 30th term is 1:3.

$:\implies\sf a_{10} : a_{30} = 1 : 3$

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

$:\implies\sf 3(a + 9d) = a + 29d$

$:\implies\sf 3a + 27d = a + 29d$

$:\implies\sf 3a - a = 29d - 27d$

$:\implies\sf 2a = 2d$

$:\implies\sf a = d\qquad\qquad\bigg\lgroup\bf eq\;(2) \bigg\rgroup$

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☯ Now, Solving eq (1) and eq (2) we get,

$:\implies\sf a = d = 2$

Now, Finding 13th term of AP,

$:\implies\sf a_{13} = a + 12d$

$:\implies\sf a_{13} = 2 + 12 \times 2$

$:\implies\sf a_{13} = 2 + 24$

$:\implies{\boxed{\frak{\purple{a_{13} = 26}}}}\;\bigstar$

$\therefore\;{\underline{\sf{Hence,\;first\;and\;13^{th}\;term\;of\;AP\;is\;2\;and\;26\; respectively.}}}$