Question

find the sum of the first 20terms of an AP inwhich 3rdterm is 7and the 7th term is two more than twice of its 3rd term

Answer 1

s₂₀ = ?
s₂₀ = n/2[2a+(n-1) d]
a₃= a+2d = 7 (given)
a₇= 2 + 2(a₃) = 2+ 2x7 = 2+14 = 16
a₇= a + 6d =16
a+2d =7.......(i)
a+6d=16.......(ii)
by elimination method ,
eqn...ii - i 
therefore , 6d - 2d = 16-7
4d= 9
d=2.25
substitute d = 2.25 in eqn ...i 
a + 2x 2.25  = 7
a+4.5 = 7 
a =2.5
so,
s₂₀=20/2 [2(2.5)+(20-1) 2.25]
s₂₀=10[5  + 42.75]
s₂₀=10(47.75)
s₂₀ = 477.5
 
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Answer 2

Given,a3=7
a7=2+a3
that is a7=9
so,
a+(n-1)d=an
a+2d=7
a+6d=9
sub. equations
we get,
4d= 2
so d=½
now, a+2d=7
a+2×½=7
so, a=7
now,
Sn = n÷2(2a+ (n-1)d)
n=20
so,
S20=20÷2(2×7+(19)×½)
=10(14+ 19×½=37÷2)
=10(18.5)=185
HOPE IT helps