2007-02-03 01:55:06 · 5 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
This is not a series. This is an equation.
There are 13 elements, subtract 2 from each (gives 2*13) and divide by 5 (gives 5(0+1+2+....+12). Next X-add in reverse order ie. 12+1, 11+2, etc which give 6 elements of the number 13 or (6*13).
2*13 + 5(0+1+2+3+4+5+6+7+8+9+10+11+12)
=26+5(13+13+13+13+13+13)
=26+5*13*6
=416
2007-02-03 02:02:58 · answer #1 · answered by Poncho Rio 4 · 0⤊ 0⤋
The last term is 62, first is 2 and common difference is 5. Hence there are n terms given by 62=2+(n-1)*5. Hence n=13.
So, the sum of the series is (n/2)*(first + last term)=13*64/2
=416.
2007-02-03 02:02:00 · answer #2 · answered by greenhorn 7 · 0⤊ 0⤋
416
there's a pattern...
2+7+12+17+22+27+32+37+42+47+52+57+62=416
2007-02-03 01:59:55 · answer #3 · answered by ...breezy...bluez.. 1 · 0⤊ 0⤋
Using t=a+(n-1)d
we get, 62=2+(n-1)5
or, (62-2)/5 = n-1
or, 12=n-1
ie, n=13.
S=n/2(2a+(n-1)d)
ie, S=13/2 (2*2+(13-1)5)
ie, S=13/2 (4+12/5)
or S=13/2 * 32/5
or, S=416/10=41.6
2007-02-03 02:04:48 · answer #4 · answered by Anindita 1 · 0⤊ 0⤋
2+7+12+......62
first term=a=2
common difference=d=5
last term=l=62
no.of terms=n=[(l-a)/d]+1=13
sum of terms=n/2(a+l)=13/2(64)=416
2007-02-03 02:58:15 · answer #5 · answered by iron muncher 3 · 0⤊ 0⤋