Question

1. Which term of the AP: 121, 117, 113,..., is
its first negative term?
[Hint: Find n for a <0]
n
2. The sum of the third and the seventh terms
of an AP is 6 and their product is 8. Find
the sum of first sixteen terms of the AP.
3. A ladder has rungs 25 cm apart.
(see Fig. 5.7). The rungs decrease
uniformly in length from 45 cm at the​

Answer 1

Answer:

(1). 32 term (2). 76

Step-by-step explanation:

1). d = - 4

$a_{n}$ = $a_{1}$ + (n - 1)d < 0

121 + (n - 1)(- 4) < 0

121 - 4n + 4 < 0

- 4n + 125 < 0

- 4n < - 125

n > (- 125) / (- 4)

n > 31.25

$a_{32}$ is the first negative number.

$a_{31}$ = 121 + (31 - 1)(- 4) = 121 - 120 = 1 > 0

$a_{32}$ = 121 + (32 - 1)(- 4) = 121 - 124 = - 3 < 0

2). $a_{3}$ × $a_{7}$ = 8

$a_{3}$ + $a_{7}$ = 6

Let $a_{3}$ = x and $a_{7}$ = y

x = 6 - y

y(6 - y) = 8

6y - y² = 8

y² - 6y + 8 = 0

(y - 4)(y - 2) = 0

$y_{1}$ = 2

$y_{2}$ = 4

$a_{3}$ = $a_{1}$ + (3 - 1)d

$a_{7}$ = $a_{1}$ + (7 - 1)d

$a_{1}$ + 2d = 2 .... (1)

$a_{1}$ + 6d = 4 .... (2)

(2) - (1)

4d = 2

d = 1/2

$a_{1}$ + 2(1/2) = 2 ===> $a_{1}$ = 1

$S_{16}$ = $\frac{n}{2}$ [2$a_{1}$ + (n-1)d]

$S_{16}$ = 8[2 + 15(1/2)] = 16 + 60 = 76

$S_{16}$ = 76

3). The task is not clear. Please re-post it with correction and image of Fig. 5.7