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ARITHMETIC PROGRESSIONS 109<br />

Here, a = 100, d = 50 and n = 21. Using the formula :<br />

S = ✁ ✂ , ✄<br />

n<br />

2<br />

2 a ( n 1) d<br />

21 21<br />

we ✆<br />

have S = ☎ 2 ✝ 100 ✞ (21 ✟ 1) ✝ 50 = ✠ 200 ✞ ✡ 1000<br />

2<br />

2<br />

= 21 1200<br />

2 ✝ = 12600<br />

So, the amount of money collected on her 21st birthday is Rs 12600.<br />

Hasn’t the use of the formula made it much easier to solve the problem?<br />

We also use S n<br />

in place of S to denote the sum of first n terms of the AP. We<br />

write S 20<br />

to denote the sum of the first 20 terms of an AP. The formula for the sum of<br />

the first n terms involves four quantities S, a, d and n. If we know any three of them,<br />

we can find the fourth.<br />

Remark : The nth term of an AP is the difference of the sum to first n terms and the<br />

sum to first (n – 1) terms of it, i.e., a n<br />

= S n<br />

– S n – 1<br />

.<br />

Let us consider some examples.<br />

Example 11 : Find the sum of the first 22 terms of the AP : 8, 3, –2, . . .<br />

Solution : Here, a = 8, d = 3 – 8 = –5, n = 22.<br />

We know that<br />

☛ S= ☞ ✞ ✟<br />

n<br />

2<br />

2 a ( n 1) d<br />

22<br />

Therefore, ✌ ✍<br />

S = 16 ✞ 21 ( ✟ 5) = 11(16 – 105) = 11(–89) = – 979<br />

2<br />

So, the sum of the first 22 terms of the AP is – 979.<br />

Example 12 : If the sum of the first 14 terms of an AP is 1050 and its first term is 10,<br />

find the 20th term.<br />

Solution : Here, S 14<br />

= 1050, n = 14, a = 10.<br />

As ✎ ✏<br />

S ✞ ✟<br />

n<br />

=<br />

✑ so, ✒<br />

1050 ✞ =<br />

n<br />

2 a<br />

2<br />

( n 1) d ,<br />

14 20<br />

2 13 d = 140 + 91d

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