Q. 55.0( 1 Vote )

# Find the sum of the first 40 terms of the series 12 – 22 + 32 – 42 + … .

Let the sum of n terms be,

Sn= 12 –22 +32 –42+52 –62+72 –82 ….

= (12 –22) + (32 –42) + (52 –62) + (72–82) …..

= (1–4) + (9–16) + (25–36) + (49–64)……..

= –3 –7 –11 –15………………. [No. of terms ]

This represents an A.P.

(NOTE: Here the number of terms has been halved. If n= no. of terms in original series and m= no. of terms in new A.P. then )

Now, here a= –3, d= (–7 – (–3)) = –4

Sum of m terms =

=

=

=

=

=

As n= 40,

Required sum =

= –820

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