find the sum of arithmetic series in which a1=5, a4=17 and n=9
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Answered by
8
first term a1 = 5
fourth term a4=17
given number of terms = 9
now,
difference d = (17-5)/3
=12/3
=4
sum = 9/2 { 2(5) +(9-1)4}
= 9/2 { 10 + 32}
= 9/2 { 42 }
= 9 × 21
= 189......answer
note ,
I used formula
{(a)n - (a)m}÷(n-m)
_-_-_-_-_ Thanks _-_-_-_-_
fourth term a4=17
given number of terms = 9
now,
difference d = (17-5)/3
=12/3
=4
sum = 9/2 { 2(5) +(9-1)4}
= 9/2 { 10 + 32}
= 9/2 { 42 }
= 9 × 21
= 189......answer
note ,
I used formula
{(a)n - (a)m}÷(n-m)
_-_-_-_-_ Thanks _-_-_-_-_
Ayush12072000:
But differences will only come if a2-a1
you are telling that difference can only find by a2-a1 & it is your wrong though , difference can be find by another formula that is{ (a)n - (a)m}/(m-n)
What is m and n
any integer
term number
Thank you
welcome
From which std you are and which board
Answered by
3
Hi !
a = 5
a₄ = 17
a + 3d = 17
5 + 3d = 17
3d = 12
d = 4
n = 9
To find : S₉
Sn = n/2 [ 2a + ( n - 1) d ]
= 9/2 [ 2 x 5 + [ 9 - 1 ] 4 ]
= 9/2 [ 10 + 8 x 4 ]
= 9/2 [ 10 + 32 ]
= 9/2 [ 42]
= 189
a = 5
a₄ = 17
a + 3d = 17
5 + 3d = 17
3d = 12
d = 4
n = 9
To find : S₉
Sn = n/2 [ 2a + ( n - 1) d ]
= 9/2 [ 2 x 5 + [ 9 - 1 ] 4 ]
= 9/2 [ 10 + 8 x 4 ]
= 9/2 [ 10 + 32 ]
= 9/2 [ 42]
= 189
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