Question

The fourth term of an Arithmetic Progression (AP) is 12 and the tenth term is 24.find the 61st term of this AP.

Answer 1

Given :

• Fourth term of an AP is 12
• Tenth term of an AP is 24

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Need To find :

• 61st term of AP?

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Solution :

• Fourth term of an AP is 12

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$\star\;{\boxed{\sf{a_4 = a + 3d}}}$

$:\implies\sf 12 = a + 3d$

$:\implies\sf a = 12 - 3d\qquad\qquad\bigg\lgroup\bf eq\;(1) \bigg\rgroup$

Also,

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• Tenth term of an AP is 24

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$\star\;{\boxed{\sf{a_{10} = a + 9d}}}$

$:\implies\sf 24 = a + 9d\qquad\qquad\bigg\lgroup\bf eq\;(2) \bigg\rgroup$

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☯ Putting eq (1) in eq (2),

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$:\implies\sf 24 = (12 - 3d) + 9d$

$:\implies\sf 24 = 12 - 3d + 9d$

$:\implies\sf 24 - 12 = 6d$

$:\implies\sf 12 = 6d$

$:\implies\sf d = {\dfrac{12}{6}}$

$:\implies{\boxed{\sf{\purple{d = 2}}}}\;\bigstar$

Now, Putting value of d in eq (1),

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$:\implies\sf a = 12 - 3(2)$

$:\implies\sf a = 12 - 6$

$:\implies{\boxed{\sf{\purple{a = 6}}}}\;\bigstar$

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☯ Now, Finding 61st term of this AP,

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$:\implies\sf a_{61} = a + 60d$

$:\implies\sf a_{61} = 6 + 60(2)$

$:\implies\sf a_{61} = 6 + 120$

$:\implies{\boxed{\sf{\pink{a_{61} = 126}}}}\;\bigstar$

$\therefore\;{\underline{\sf{Hence,\;61^{st}\;term\;of\;this\;AP\;is\; \bf{126}.}}}$

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$\qquad\boxed{\underline{\underline{\bigstar \: \bf\:Formula\:Related\:to\:AP\:\bigstar}}}$

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$\sf (i)\;The\; n^{th}\;term\;of\;an\;AP\; = \; \red{a_n + (n - 1)d}$

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$\sf (ii)\;Sum\;of\;n\;term\;of\;an\;AP\; = \; \purple{S_n = \dfrac{n}{2} \bigg\lgroup\sf 2a + (n - 1)d \bigg\rgroup}$

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$\sf (iii)\;Sum\;of\;all\;terms\;of\;AP\;having\;last\:term\;as\;'l'\; = \; \pink{ \dfrac{n}{2}(a + l)}$

Answer 2

Answer:

a61=126

Step-by-step explanation:

see the given picture to understand the solution