in alphabetic sequence the sum of first 10 terms is 230 and the sum of first 16 term is 560 find the first term and common difference of the sequence is algebraic expression of the sum
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S10=230
S16=560
10a+45d=230→1
560=16a+120d→2
1÷5&2÷8
2a+9d=46→3
2a+15d=70→4
4-3
3d=24
d=8
Substitute the value of d in 3
∴a=-13
S16=560
10a+45d=230→1
560=16a+120d→2
1÷5&2÷8
2a+9d=46→3
2a+15d=70→4
4-3
3d=24
d=8
Substitute the value of d in 3
∴a=-13
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