Math, asked by kinghacker, 1 day ago

Find the sum of the following by short-cut method : (a) 2096 + 9 (b)4816 + 9 (c) 1964 + 99 (d) 3178 + 99 (e) 4196 + 995 (f) 30968 + 9996​

Answers

Answered by crankybirds31
2

Answer:

"=49*25= 1225.

Alternatively, you can directly calculate by Sn =(first term+last term) * no of term in progression/2. no of term can calculate by (last term-first term)/d +1 or alternatively you can calculate no of term by, greatest integer of ((last term - first term )/d), where d is term difference.

Sn=(1+49)*49/2= 1225. "

Answered by crankybirds31
2

Answer:

"=49*25= 1225.

Alternatively, you can directly calculate by Sn =(first term+last term) * no of term in progression/2. no of term can calculate by (last term-first term)/d +1 or alternatively you can calculate no of term by, greatest integer of ((last term - first term )/d), where d is term difference.

Sn=(1+49)*49/2= 1225. "

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