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The sum of the first 45 terms of an A.P....

The sum of the first 45 terms of an A.P. is 3195. Complete the following activity to find the 23rd term.

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`S_(n) =S_(450 = 3195`
Let the first term of the A.P. be a and the common difference d.
`S_(n) = ( n )/(2) [ 2a +square ]` ….(Formula)
`:. S_(45) = ( 45)/(2) [ 2a + square ]` …(Substituting the values )
`:. 3195 = ( 45)/(2) [ 2a + 44d]`
`:. 3195 = 45 xx square `
`:. a + 22d = square `
`:. a + 22d = 71` ...(1)
Now, 23 rd term is `t_(23)`.
`t_(n) = square ` .....(Formula)
` :. t_(23) = a+ ( 23-1) d `
`:. t_(23) = a+ 22d `
`:. t_(23) = square ` ....[From (1) ]
Activity `: S_(n) = ( n )/(2)[2a + ( n-1) d ]`
`:. S_(45) /(2) [2a+ ( 45-1) d]`
`:. 3195 = ( 45)/(2) [ 2a+ 44d]`
`:. 3195 = 45 xx a + 22d`
`:. a + 22d = ( 3195)/(45)`
`:. a+ 22d = 71` ...(1)
Now, 23rd term is `t_(23)`.
`t_(n) = a+(n-1)d ` ...(Formula)
`:. t_(23) = a+ (23-1)d`
`:. t_(23) =a+22d`
`:. t_(23) = 71` ...[From (1)]
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