pls solve the question in the picture (chapter is arithmetic progression)
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993
when multiplied by 5 its remained only 3.
when multiplied by 5 its remained only 3.
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We know that an = a + (n - 1) * d
998 = 103 + (n - 1) * 5
998 - 103 = (n - 1) * 5
895/5 = n - 1
n = 180.
We know that sn = n/2(a+I)
= (180/2)(103 + 998)
= 99090.
Sum of 3-digit numbers = 99090.
998 = 103 + (n - 1) * 5
998 - 103 = (n - 1) * 5
895/5 = n - 1
n = 180.
We know that sn = n/2(a+I)
= (180/2)(103 + 998)
= 99090.
Sum of 3-digit numbers = 99090.
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