Question

a)Fifth term of an arithmetic sequence is 40 and the tenth term is 20 what is the fifteenth term of this sequence

b)How many termsvof this sequence will be added to get zero as the sum?

Answer this question with step by step explanation and I will mark as Brainlist​

Answer 1

Answer:

let a & d be the first term and the common difference respectively

ATQ,

$t_{5} = a + (n - 1)d$

$40= a + (5 - 1)d$

$40= a + 4d$

$40 - 4d = a$            .......... (equation 1)

$t_{10} = a + (n - 1)d$

$20= a + (10 - 1)d$

$20= a + 9d$

$20= 40 - 4d + 9d$           (from equation 1)

$20 - 40= 5d$

$-20 = 5d$

$-4 = d$

putting the value of d in equation 1

a = 40 - 4d

a = 40 - 4(-4)

a = 40 + 16

a = 46

$t_{15} = a + (n - 1)d$

$t_{15}= 46 + (15 - 1)(-4)$

$t_{15}= 46 + 14(-4)$

$t_{15} = 46 - 56$

$t_{15} = - 10$

b) $t_{n} = 0$

$t_{n} = a + (n - 1)d$

$0 = 46 + (n - 1)(-4)$

$46 = (n - 1)(-4)$

$\frac{-46}{-4} = (n - 1)$

$\frac{23}{2} = (n - 1)$

$\frac{23 + 2}{2} = n$

$\frac{25}{2} = n$

since n is in fraction thus there are no terms in the AP which will add up to zero as the sum

Answer 2

$\large\bf\underline {To \: find:-}$

• we need to find the fifteenth term of AP
• How many terms of this sequence will be added to get zero as the sum?

$\huge\bf\underline{Solution:-}$

$\bf\underline{\red{Given:-}}$

• 5th term = 40
• 10th term = 20

we know that,

$\bigstar \star \red{ \bf \: a_n = a + (n -1)d}$

So,

• 5th term = a + 4d = 40 .....1)
• 10th term = a + 9d = 20 ......2)

⇝ From 1) and 2)

⠀⠀⠀ a + 4d = 40

⠀⠀⠀ a + 9d = 20

⠀⠀⠀ -- ⠀-- ⠀⠀-- ⠀⠀⠀

⠀⠀⠀⠀- 5d = 20

⠀⠀⠀⠀⠀⠀d = -4

• ▶ putting d = -4 in 1)

↛a + 4d = 40

↛ a + 4 × -4 = 40

↛ a = 40 + 16

↛ a = 56

So,

• First term of AP (a) = 56
• common difference (d) = -4

Now,

• Fifteenth term of AP = a + 14d

↛ a15 = a + 14d

↛ a15 = 56 + 14 × -4

↛ a15 = 56 - 56

↛ a15 = 0

So,

• Fifteenth term of AP is 0

Now,

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b)How many termsvof this sequence will be added to get zero as the sum?

we know that,

$\bigstar \star \red{ \tt \: S_n = \frac{n}{2} [2a + (n -1 ) d]}$

$\tt \leadsto \:0= \frac{n}{2} \bigg\{2 \times 56 + (n - 1) \times( - 4 )\bigg \}$

⠀

$\tt \leadsto \:0= \frac{n}{2} \bigg \{112 - 4n + 4 \bigg \}$

$\tt \leadsto \:0=n \bigg \{116 - 4n \bigg \}$

$\tt \leadsto \:0=116n - 4 {n}^{2}$

$\tt \leadsto \:4n {}^{2} - 116n = 0$

$\tt \leadsto \:4n(n - 29) = 0$

$\tt \leadsto \:n - 29 = \frac{0}{4n}$

$\tt \leadsto \:n = 29$

Hence,

• Sum of 29 terms will be 0.

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