Question

139. (112 + 122 + 132 + ... + 202) = ?​

Answer 1

Answer:

1570

Step-by-step explanation:

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This is an AP,

where

a(first term)=112

d(common difference)= 10

l or Tn(Last term)=202

First, we have to find number of terms in AP,

Tn = a + (n-1)d

202 = 112 + (n-1)10

90 = (n-1)10

n-1 = 9

n= 10

Now, sum of AP

= n/2 * [ 2a + (n-1)d]

= 10/2 * [ 2*112 + (10-1)10 ]

=5 * [224+90]

=5 * 314

= 1570