Question

Sum of first 55 terms in an A.P. is 3300, find its 20th term.​

Answer 1

Answer:

$it \: is \: given \: that \\ n = 55 \\ ap = 3300 \\ \\ we \: know \: that \\ {s}^{55} = \frac{55}{2} (2a + 55 - 1)d \\ 3300 = \frac{55}{2} (2a + 54d) \\ 3300 = 55(a + 27d) \\ \frac{3300}{55} = a + 27d \\ a + 27d = 60..1equation \\ now \: n = 28 \\ \\ t = a + (n - 1)d \\ t = a +( 28 - 1)d \\ t = a + 27d \\ t = 60 \\ \\ twenty \: eight \: term \: of \: ap \: is \: 60 \\ hence \: 28 \: term \: is \: 60$

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Answer 2

Correct Question:

• Sum of first 55 terms of an AP is 3300, find it's 28th term.

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Given:

• Sum of first 55 terms, (S₅₅) of AP is 3300

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To find:

• 20th term of AP (a₂₀)

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Solution:

• We have formula for sum of nth terms of AP::

» Sₙ = n/2 [ { 2a + ( n-1 } d ]

:: Substitute the value of n=55

» S₅₅ = 55/2 [ { 2a + ( 55-1 ) d } ]

:: Put value of S₅₅=3300

» 3300 = 55/2 [ { 2a + 54d } ]

:: Now transport 55/2 from RHS to LHS

[ It will change into its recipocated form after transportation]

» 3300 × 2/55 = 2a + 54d

:: Now divide both sides by 2

» [ 3300 × (2/55) ] ÷ 2 = ( 2a + 54d ) ÷ 2

:: Now solving it

» 3300/55 = a + 27d

» 60 = a + 27d

» 60 = a₂₈  

Hence the required 28th term of AP is 60.

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