Question

In an ap 3+9+15+21+_ _ _ _ find a, CD,an,a20 s25, s 20 - s19
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Answer 1

The given series 3+9+25+21+....

let the given series be a1+ a2+ a3+ a4+...

therefore first term a = a1 =3

Common difference d = a2-a1=a3-a2

a2-a1 = 9-3 =6

a3-a2 = 15-9 = 6

therefore d = 6

• an=a + (n-1)d
• sn =(n/2)(2a+(n-1)d)

a20 = 3+(20-1)6

=3+114

=117

then for s 25, s20 - s19 u can substute in the forma of sn and get the answer...

Answer 2

Step-by-step explanation:

3+9+15+21+_ _ _ _

first term (a^1) = 3

common difference(d) = a2 - a1

= 9 - 3 = 6

an = a+(n-1)d

= 3+(n-1)6

=3+6n-6

= 6n - 3

a20 = a1 + 19 (d)

= 3 + 19 (6)

= 3 + 96

= 99

s25 = n/2 [2a+(n-1)d]

= 25/2 [ 2(3)+(25-1)6]

= 25/2 [6+(24)6]

= 25/2 [6+144]

=25/2[150]

=25(75)

=1874

s20 = 20/2 [ 2(3)+(20-1)6]

= 10 [ 6+(19)6]

= 10 [6+96]

=10(114)

= 1140

s19 = 19/2 [2(3)+(19-1)6]

= 19/2 [ 6+(18)6]

= 19/2 [6+108]

= 19/2[ 114]

=19[57]

=1083

s20 - s19= 1140 - 1083

= 57

I hope it may be help you..........

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