hey thanks alot!!!!:D i understand already!
this one quite weird qn, can help me explain the question. i
don't understand the qn. btw is the 2002 a lvl paper 1 qn 10.
an ap in which the common diff is twice the 1st term is called a
special ap Show that the sum of the first n terms of a special ap
with 1st term a is n(square)a.
The (M-2)th term of a special ap is 663 and the Mth term is 731.
Find M and find also the sum of the first M terms.
The sum of the first N terms of another special ap is S. FInd in
the terms of S, the sum of the next N terms.
Answer: 22,8228, 35
Let first term = a, common difference = 2a
Sn = (n/2)*(2a + (n-1)(2a))
= (n/2)*(2a + 2an - 2a)
= n²a (shown)
Mth term - (M-2)th term = 2 * common difference
731 - 663 = 2 * 2a
a = 17, d = 34
731 = 17 + (M-1)(34)
M = 22
Sum of first M terms =22² * 17 = 8228
Sn = n²a
S2n = 4n²a
Difference of next N terms = S2n - Sn
= 3n²a
= 3S ===> (not 35 hor)
