Question

The seventh term of an arithmetic series is 6 and the eighteenth term is 22.5 Calculate (a) The common difference of the series. (b) The first term of the series. Given also that the sum of the first n terms of the series is 252, (c) Find the value of n. ​

Answer 1

Step-by-step explanation:

A) a7=6

a18=22.5

a+6d=6

a+17d=22.5

subract eq 1 and 2

will get d=1.5

Answer 2

STEP-BY-STEP EXPLANATION:

.

$:\implies \bf {a}_{n} = a + (n - 1)d$

$:\implies\tt {a}_{7} = 6 \: \: \: ...(given)$

$:\implies\tt {a}_{7} = a + (7 - 1)d$

$:\implies\tt 6= a + 6d \: \: \: ...(1)$

$:\implies\tt {a}_{18} = 22.5 \: \: \: ...(given)$

$:\implies\tt {a}_{18} = a + (18 - 1)d$

$:\implies\tt 22.5 = a + 17d \: \: \: ...(2)$

$\bf Substract \: Eq[1] \: from \: Eq[2],$

$:\implies\tt 22.5 - 6 = (a + 17d) - (a + 6d)$

$:\implies\tt 16.5 = a +17d - a - 6d$

$:\implies\tt 16.5 = 11d$

$:\implies\bf d = 1.5$

$\bf Substituting \: this \: value \: of \: d \: in \: Eq [1],$

$:\implies\tt 6= a + 6d \: \: \: ...(1)$

$:\implies\tt 6 = a + 6(1.5)$

$:\implies\tt 6 = a + 9$

$:\implies\bf a = - 3$

$:\implies\bf {S}_{n} = \frac{n}{2} [2a + (n - 1)d]$

$:\implies\tt 252 = \frac{n}{2} (2( - 3) + (n - 1)(1.5))$

$:\implies\tt 252 = \frac{n}{2} ( - 6 + \frac{3}{2} n - 1.5)$

$:\implies\tt 252 = \frac{n}{2} ( - 7.5 + \frac{3}{2} n)$

$:\implies\tt 252 = \frac{ - 7.5 }{2} n + \frac{3}{4} {n}^{2}$

$:\implies\tt252 = \frac{3}{4} {n}^{2} - \frac{15}{4} n$

$:\implies\tt 1008 = 3 {n}^{2} - 15n$

$:\implies\tt 1008 = 3( {n}^{2} - 5n)$

$:\implies\tt 336 = {n}^{2} -5n$

$:\implies\tt {n}^{2} = 336 + 5n$

$:\implies\tt n = \sqrt{336 + 5n}$

$\bf By \: Hit \: and \: Trial \: Method \: We \: Find,$

$:\implies\bf n = 21$

REQUIRED ANSWER,

.

• The common difference of the series is 1.5.
• The first term is -3.
• The value of n is 21.