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# HridayAg0102
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Answer 1

(19)
Sn=3n2+13n2

an=Sn−Sn−1

a25=S25−S24

a25=3(25)2+13(25)2−3(24)2+13(24)2

a25=12[3(252−242)+13(25−24)]

a25=12[3(49)+13]

a25=80
or

Let a be the first term and d the common difference of the given A.P. Then, the sums of m and n terms are given by

Sm = (m2) [2a + (m – 1) d], and

Sn = (n2) [2a + (n – 1) d] 

SmSn = m2n2

m2[2a+(m−1)d]n2[2a+(n−1)d] = m2n2

⇒ 2a+(m−1)d2a+(n−1)d = mn

⇒[2a+(m−1)d]n=[2a+(n−1)d]m

⇒2a(n−m)=d((n−1)m−(m−1)n)

⇒2a(n−m)=d(n−m)

⇒d=2a

TmTn = a+(m−1)da+(n−1)d=a+(m−1)da+(n−1)d=2m−12n−1

Answer 2

Hey there!

$S_n$ = 3n²/2 + 13n/2

First term, a = 8
Sum of two terms = 19

2nd term = 19-8 = 11
Thus,
Common difference, d = 11-3 = 8

∴ 25th term = (8 + (25 -1)×3) = 80

Alternative method:

nth term of an A.P. = (Sum of nth term ) - [Sum of (n-1)th term] 

25th term =  (Sum of 25 terms) - ( Sum of 24 terms)
                 
                 =(3×(25)²/2 +(13×25)/2) - ( 3×(24)²/2 + (13×24)/2)
                     
                 = 1100 - 1020
                     
                 = 80

Hence, the 25th Term is = 80

HOPE IT HELPS ^_^