In an A-P- the sum of first n terms is 3n2-2 - 13n-2- Find the 25th term- -3 MARKS-

Formula: 1 Mark Application: 1 Mark Solution: 1 Mark Given, S_n=3n^2+13n/2...(1) a_n=S_n S_n 1 ⇒ a_25=S_25 S_24 a_25=3(25)^2 + 13(25)/2 3(24)^2 + 13(24) ...

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Standard XII

Mathematics

Sum of n Terms

In an A.P. th...

Question

In an A.P. the sum of first n terms is $\frac{3 n^{2}}{2} + \frac{13 n}{2}$ . Find the 25th term. [3 MARKS]

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Solution

Formula: 1 Mark
Application: 1 Mark
Solution: 1 Mark

$G i v e n, S_{n} = \frac{3 n^{2} + 13 n}{2} . . . (1)$

$a_{n} = S_{n} - S_{n - 1}$

$\Rightarrow a_{25} = S_{25} - S_{24}$

$a_{25} = \frac{3 (25)^{2} + 13 (25)}{2} - \frac{3 (24)^{2} + 13 (24)}{2} [F r o m (1)]$

$a_{25} = \frac{1}{2} [3 (25^{2} - 24^{2}) + 13 (25 - 24)]$
$a_{25} = \frac{1}{2} [3 (49) + 13]$
$a_{25} = 80$

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In an A.P. the sum of first n terms is 3n22+13n2. Find the 25th term. [3 MARKS]

Formula: 1 Mark Application: 1 Mark Solution: 1 Mark Given, Sn=3n2+13n2...(1) an=Sn−Sn−1 ⇒a25=S25−S24 a25=3(25)2+13(25)2−3(24)2+13(24)2 [From(1)] a25=12[3(252−242)+13(25−24)] a25=12[3(49)+13] a25=80

In an A.P. the sum of first n terms is $\frac{3 n^{2}}{2} + \frac{13 n}{2}$ . Find the 25th term. [3 MARKS]