Question

Find the sum of first 20 terms of an ap in which the 3rd term is 7 and 7th term is two more than the thrice of its 3rd term

Answer 1

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Answer 2

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

Here,

• 3rd term of AP is 7
• 7th term is two more than the thrice of its 3rd term

${\underline{\frak{We\;have}}} \begin{cases} & \sf a_3 = 7 \\ & \sf a_7 = 3a_3 + 2 \end{cases}$

We have to find, The sum of 20 terms of an A.P.

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$\quad\qquad\star\;{\boxed{\sf{\purple{a_3 = 7}}}}$

$\: \: \: :\implies\sf a + 2d = 7$

$\: \: \: :\implies\sf a = 7 - 2d\qquad\qquad\bigg\lgroup\bf eq\:(1)\bigg\rgroup$

Now,

$\quad\qquad\star\;{\boxed{\sf{\purple{a_7 = 3a_3 + 2}}}}$

$:\implies\sf a + 6d = 3[a + 2d] + 2$

$\: \: :\implies\sf a + 6d = 3a + 6d + 2$

$\: \: :\implies\sf a + \cancel{6d} = 3a + \cancel{6d} + 2$

$\qquad \quad:\implies\sf a = 3a + 2$

$\qquad \quad:\implies\sf a - 3a = 2$

$\qquad \quad \: \: :\implies\sf - 2a = 2$

$\qquad \quad \: :\implies\sf a = - \cancel{ \dfrac{2}{2}}$

$\qquad \quad :\implies{\boxed{\frak{\pink{a = -1}}}}\; \bigstar$

$\star\;{\underline{\frak {Putting\;value\;of\;a\;in\;eq\;(1),}}}$

$\qquad \: \: \: \: \: :\implies\sf a = 7 - 2d$

$\qquad \: \: \: :\implies\sf -1 = 7 - 2d$

$\qquad:\implies\sf -1 - 7 = - 2d$

$\qquad \: \: \: \: \: \: :\implies\sf - 8 = - 2d$

$\qquad \: \: \: \: \: \: \: :\implies\sf d = \cancel{ \dfrac{-8}{-2}}$

$\qquad \: \: \: \: \: \: \: :\implies{\boxed{\frak{\pink{d = 4}}}}\; \bigstar$

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☯ We have to find, sum of 20th term of A.P. So,

$\dag\;{\underline{\frak{We\;know\;that,}}}$

$\qquad \quad\star\;{\boxed{\sf{\purple{S_n = a + \dfrac{n}{2} \bigg\lgroup\sf 2a + (n - 1)d \bigg\rgroup}}}}$

$:\implies\sf S_{20} = a + \dfrac{20}{2} \bigg\lgroup\sf 2 \times -1 + (20 - 1) \times 4 \bigg\rgroup$

$\qquad \quad \: \: :\implies\sf S_{20} = 10 \bigg\lgroup\sf - 2 + 19 \times 4 \bigg\rgroup$

$\qquad \quad \: \: \: \: \: \: \: :\implies\sf S_{20} = 10 \bigg\lgroup\sf - 2 + 76 \bigg\rgroup$

$\quad \quad \qquad \qquad:\implies\sf S_{20} = 10 \bigg\lgroup\sf 74 \bigg\rgroup$

$\quad \quad \qquad \qquad \:\: :\implies{\boxed{\frak{\pink{S_{20} = 740}}}}\; \bigstar$

$\therefore\;{\underline{\sf{20th\;term\;of\;A.P.\;is\;\bf{740}.}}}$