[tex]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[/tex]
Correct Question:
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Given:
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To find:
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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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Answer:
[tex]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[/tex]
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