if the sum of the 12th and 22nd terms of an ap is 100 then the sum of the first 33 terms of the ap is
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nth Term of an Arithmetic Progression, Tn=a+(n−1)d
Sum of n terms, Sn=(n/2)∗[2a+(n−1)∗d
Where, a = first term of Progression, d = common difference (Second term - first term or Third - second term etc.)
12th Term, T12 = a+(12−1)∗d = a+11d
22th Term, T22=a+(22−1)∗d = a+21d
Given that
(a+11d) + (a+21d)=100
2a + 32d=100
Sum of 33 terms,
S33=(33/2)∗[2a+(33−1)∗d
=(33/2)∗[2a+32d]
=(33/2)∗100
=1650
Sum of n terms, Sn=(n/2)∗[2a+(n−1)∗d
Where, a = first term of Progression, d = common difference (Second term - first term or Third - second term etc.)
12th Term, T12 = a+(12−1)∗d = a+11d
22th Term, T22=a+(22−1)∗d = a+21d
Given that
(a+11d) + (a+21d)=100
2a + 32d=100
Sum of 33 terms,
S33=(33/2)∗[2a+(33−1)∗d
=(33/2)∗[2a+32d]
=(33/2)∗100
=1650
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