The first five terms of a linear sequence are -3, 4, 11, 18, 25.
a Find the 100th term of the sequence.
b Show that 102 is a member of the sequence and find its term number.
please send the way too dont just give the answer
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Answer:
Here is ur answer...
Step-by-step explanation:
sequence is -3,4,11,18,25
a) common difference = x2 - x1
=4 -- 3
= 4 + 3
= 7
100th term = x1 +99d
= -3 +99 × 7
= -3 + 693
= 690
b) xn - x1/d
= 102 --3/7
=102+3/7
= 105/7 =15
Difference of terms is the multiple of common difference
there fore, 102 is a term of this sequence
15 is its position..
Hope it helps u....
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