Question 1 (e)

Find:

23 – (–12)

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Solution

23 – (–12) = 23 + 12 = 35

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Find the sum of the series $\frac{1^{3}}{1} + \frac{1^{3} + 2^{3}}{1 + 3} + \frac{1^{3} + 2^{3} + 3^{3}}{1 + 3 + 5} + - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 12 t e r m s$

Q. Find the sum of first

10

terms of following series

S = \frac{3 (1)}{1} + \frac{5 (1^{3} + 2^{3})}{1^{2} + 2^{2}} + \frac{7 (1^{3} + 2^{3} + 3^{3})}{1^{2} + 2^{2} + 3^{2}} + . . . .

Find:

(a) 35 − (20) (b) 72 − (90)

(e) 23 − (− 12) (f) (− 32) − (− 40)

Q. Find:
(a) 35 - (20)
(b) 72 - (92)
(c) (-15) - (-18)
(d) (-20) -(-13)
(e) 23 - (-12)
(f) (-32) - (-40)

Q. Find the 11th term of the AP 1, 12, 23, 34, ...

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Subtraction of Integers

Standard VI Mathematics

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Question 1 (e) Find: 23 – (–12)

23 – (–12) = 23 + 12 = 35

Question 1 (e)

Find:

23 – (–12)