The third and seventh terms of an ap are respectively -2 and 10 find the value of the first term and number of term which must be add to give sum of 410
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Tn = a + (n-1) d
so for example T3 = a + 2 d and T7 = a + 6 d
that gives you two linear equations to solve for a and d
then use the equation for sum of terms to find the n for sum of 410
so for example T3 = a + 2 d and T7 = a + 6 d
that gives you two linear equations to solve for a and d
then use the equation for sum of terms to find the n for sum of 410
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Math is Fun arithmetic sequence
Math is Fun arithmetic sequence
I am not about to do it for you. It is pure plug and chug.
a+2d = -2
a+6d = 10
solve for a and d, and then you have
n/2 (2a+(n-1)d) = 410
Since you now know a and d, just use them to solve for n
a+6d = 10
solve for a and d, and then you have
n/2 (2a+(n-1)d) = 410
Since you now know a and d, just use them to solve for n
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