# arithmetic progression

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the 9th term of an AP is 52 while the 12th term is 70. find the sum of it's 20th term?

• arithmetic progression -

9th term = 52 ---> a + 8d = 52
12th term = 70 --> a + 11d = 70
subtract them
3d = 18
d = 6
then a+48 = 52
a = 4

20th term = a + 19d = 4 + 114 = 118

• arithmetic progression -

118

• arithmetic progression -

118

• arithmetic progression -

9th term = 52
a+8d=52
12th term = 70
a+11d =70
Subtract them
a+8d=52
-
a+11d =70
3d=18
d=6 buy dividing both side by 3
To fine a
a+8d=52
When d= 6
a+8(6)=52
a+48=52
a=52-48
a=4
Then the sum of 20th term
Will be
S20= n/2[2a + (n-1)d]
S20= 20/2[2(4) + (20-1)6]
S20=10[8+19(6)]
S20=10[8+114]
S20=10[122]

• arithmetic progression -

9th term = 52
a+8d=52
12th term = 70
a+11d =70
Subtract them
a+8d=52
-
a+11d =70
3d=18
d=6 buy dividing both side by 3
To fine a
a+8d=52
When d= 6
a+8(6)=52
a+48=52
a=52-48
a=4
Then the sum of 20th term
Will be
S20= n/2[2a + (n-1)d]
S20= 20/2[2(4) + (20-1)6]
S20=10[8+19(6)]
S20=10[8+114]
S20=10[122]

• arithmetic progression -

Since is sum of 20th term not the 20th term then the answer is 1220

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