Question

for an A.P find S7 and t20 if a = -5 and d=4
plzz solve this question fast i will mark you as brainlist

Answer 1

solution
we know that
tn = a + (n-1)d
where,
tn is the nth term
a is the first term
n is the number of terms
d is the common difference
so,
t20= a+19d

= -5+(19×4)
= 71
also we know
sum of n terms of an ap
sn = n/2(2a+(n-1)d)
so
s7 = 7/2(2a+(6d))
= 3.5(-10+ 24)
= 49

hope that helps
please mark my answer as brainliest

Answer 2

Hey there !

Solution:

Given: ( a ) = -5, ( d ) = 4

To find: S₇ and t₂₀

Proof:

=> tₓ = a + ( x - 1 ) d

Here x refers to the number of terms.

=> t₂₀ = a + ( 20 - 1 ) d

=> t₂₀ = -5 + ( 19 ) 4

=> t₂₀ = -5 + 76

=> t₂₀ = 71

Now lets find S₇,

Sₓ = ( x / 2 ) [ 2a + ( x - 1 ) d ]

=> S₇ = ( 7 / 2 ) [ 2 ( -5 ) + ( 7 - 1 ) 4 ]

=> S₇ = 3.5 [ -10 + ( 6 ) 4 ]

=> S₇ = 3.5 [ -10 + 24 ]

=> S₇ = 3.5 [ 14 ]

=> S₇ = 49

Hence t₂₀ = 71 and S₇ = 49.

Hope my answer helped !