Question

In an ap the sum of first n term is 3n2/2+5n/2 find a25

Answer 1

Sn=3n²/2+5n/2. ------->1
required to find:-
a25 th term

substitute
n= 1 in equation 1

S1= 3(1)²/2+5(1)/2
S1= 3/2+5/2
S1 = 4

substitute n= 2 in equation 2
S2= 3(2)²/2+5(2)/2=(3×4)/2+(10)/2 =6+5=11

formula:-
an=a1+(n-1)d

Therefore 1st term a1=. 4 ; 2nd term a2= S2-S1= 11-4=7. and d= 7-4=3

Now 25th term

an=a1+(n-1)d
here n= 25

a25= 4+(25-1)×3
= 4+24×3
a25=76.

the value of a25 = 76.

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Answer 2

Answer:

a25 = 76

Step-by-step explanation:

Given: Sn = 3n^2/2 + 5n/2

To Find: a25

Solution: Sn = 3n^2/2 + 5n/2 ----eq(1)

• put n=1 in eq(1)

S1 = 3(1)^2/2 + 5(1)/2

S1 = 3/2 + 5/2

S1 = 8/2

S1 = 4

• put n=2 in eq(1)

S2 = 3(2)^2/2 + 5(2)/2

S2 = 12/2 + 10/2

S2 = 6 + 5

S2 = 11

• a = S1 = 4

• a2 = S2-S1

a2 = 11-4

a2 = 7

• d = a2 - a

d = 7-4

d = 3

☆ a25 = a + 24d

a25 = 4 + 24(3)

a25 = 4 + 72

a25 = 76

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