Solve the equation:1+4+7+10+...+x=287
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Answered by
9
first calculate last term
last term=first term+difference of terms(no of terms-1)
x=1+3(n-1)
x=3n-2
then sum of terms=n/2(first term+last term)
287=n/2(1+3n-2)
here n=14
so last term=3n-2=40
last term=first term+difference of terms(no of terms-1)
x=1+3(n-1)
x=3n-2
then sum of terms=n/2(first term+last term)
287=n/2(1+3n-2)
here n=14
so last term=3n-2=40
Answered by
18
Sn = n/2[2a + (n-1)d]
287 = n/2[2(1) + (n-1)3]
287 x 2 = n[2 + 3n - 3]
574 = n[3n - 1]
574 = 3n² - n
3n² - n - 574 = 0
Using this formula -->
ß =( -b ± √b² - 4ac)/2a
n = 84/6
n = 14
So , n = 14th term
an = a + (n-1)d
= 1 + (14-1)3
= 1 + 13x3
= 40
[ x = 40 ]
287 = n/2[2(1) + (n-1)3]
287 x 2 = n[2 + 3n - 3]
574 = n[3n - 1]
574 = 3n² - n
3n² - n - 574 = 0
Using this formula -->
ß =( -b ± √b² - 4ac)/2a
n = 84/6
n = 14
So , n = 14th term
an = a + (n-1)d
= 1 + (14-1)3
= 1 + 13x3
= 40
[ x = 40 ]
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