The sum of first n terms of an AP is given by sn = 3n2 + 4. find its tenth term
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Answered by
1
Given Sn = 3n2 + 4n
S1 = 3×1 + 4×1 = 7.
S2 = 3×4 + 4×2 = 12 + 8 = 20.
Therfore First term a1 = 7 & Second term a2 = 20 - 7 = 13 and d = 13 - 7 = 6.
Now 25th term an = 7 + ( 25 - 1) 6
a25 = 7 + 24 x 6
a25 = 151.
S1 = 3×1 + 4×1 = 7.
S2 = 3×4 + 4×2 = 12 + 8 = 20.
Therfore First term a1 = 7 & Second term a2 = 20 - 7 = 13 and d = 13 - 7 = 6.
Now 25th term an = 7 + ( 25 - 1) 6
a25 = 7 + 24 x 6
a25 = 151.
Answered by
2
Sn=3n²+4
S1= 3x(1)²+4
=7
S2=3x(2)²+4
=48
S2=a1 +a2
48=7+a2
a2=48-7
=41
d=a2-a1
=41-7
=34
a10=a+9d
=7+9x34
=7+306
=313
S1= 3x(1)²+4
=7
S2=3x(2)²+4
=48
S2=a1 +a2
48=7+a2
a2=48-7
=41
d=a2-a1
=41-7
=34
a10=a+9d
=7+9x34
=7+306
=313
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