Question

3. For an AP S4 - S3 = 3d.
true or false​

Answer 1

The given statement is false.

Step-by-step explanation:

Let a, d be the first term and common difference respectively.

To check,  for an AP $S_4 - S_3$ = 3d,  true or false​.

We know that,

The sum of nth terms of an AP,

$S_{n} =\dfrac{n}{2}[2a+(n-1)d]$

∴ $S_{4} =\dfrac{4}{2}[2a+(4-1)d]=\dfrac{4}{2}(2a+3d)$ and

$S_{3} =\dfrac{3}{2}[2a+(3-1)d]=\dfrac{3}{2}(2a+2d)$

∴ $S_4 - S_3$

= $\dfrac{4}{2}(2a+3d)-\dfrac{3}{2}(2a+2d)$

= $\dfrac{1}{2}[4(2a+3d)-3(2a+2d)]$

= $\dfrac{1}{2}[8a+12d-6a-6d]$

= $\dfrac{1}{2}(2a+6d)$

= $\dfrac{1}{2}\times 2(a+3d)$

= a + 3d

≠ 3d, the given statement is false.

Thus, the given statement is false.

Answer 2

Step-by-step explanation:

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