Question

Find the ratio of the sum of the first 43 terms of an AP to the common difference if the sum of the first 15 terms and that of the first 27 term of this AP is equal?s

Answer 1

The required ratio is 43/2

Step-by-step explanation:

let the AP be a, a+d, a+2d..a+(n-1)d

here the last term l = 43

therefore n= 43

then sum of first 43 terms is

$S_n =\frac{n}{2}(2a+(n-1)d)$

given that $S_1_5 = S_2_7$

thus

$\frac{15}{2}(2a+14d)= \frac{27}{2}(2a+26d)\\\\\text {on solving}\\\\5a+35d= 9a+117d\\\\2a+41d=0\\\\S_4_3= \frac{43}{2}(2a+42d)\\\\S_4_3=\frac{43}{2}(2a+41d+d)= \frac{43}{2}(0+d)\\\\S_4_3= \frac{43d}{2}\\\\\frac{S_4_3}{d}=\frac{43}{2}$

hence , The ratio is 43/2

#Learn more:

If the 21st term of ap is 43 find the sum of its first 41 terms

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