Question

THE 5TH TERM OF AN AP IS 1 WHWERE AS ITS 31ST TERM IS-77 WHICH OF ITS TERM IS -17

Answer 1

Answer:

Answer:

5th term = t5 =1

31st term = t31 = 77

t31 = a+ (n-1)d

t31 = a + 30d

77 =  a +30d

a =  77 -30d ______________ (1)

t5 = a +4d

1 = a +4d

a = 4d -1  ____________(2)

Equating (1) and (2)

77 - 30d = 4d - 1

77+1 = 30 + 4d

78 = 34 d

d = 2.294

d= 2.3

a = 4d-1

a = 9.2 -1

a = 8.2

t17 = a + (17 -1) * 2.3

t17 = 8.2 + 36.8

t17 = 45 .00

Answer 2

$\huge\sf\pink{Answer}$

☞ 11th term of the AP

$\rule{110}1$

$\huge\sf\blue{Given}$

✭ In an A.P,

◕ 5th term is 1

◕ 31st term is -77

$\rule{110}1$

$\huge\sf\gray{To \:Find}$

◈ N tg term with -17 as it's value?

$\rule{110}1$

$\huge\sf\purple{Steps}$

Assume the first term of the AP to be a and it's Common Difference as d

$\bullet\underline{\textsf{\: As Per the Question}}$

$\sf 5th term = a+4d = 1 \qquad-eq(1)$

$\sf 31st term = a+30d = -77 \qquad -eq(2)$

Subtract eq(1) from eq(2)

➢ $\sf (a+30d)-(a+4d) = -77-1$

➢ $\sf a+30d-a-4d = -78$

➢ $\sf 26d = -78$

➢ $\sf d = -78/26$

➢ $\sf\red{d = -3}$

Substituting value of d in eq(1)

➝ $\sf a+4(-3) = 1$

➝ $\sf a-12 = 1$

➝ $\sf a = 1+12$

➝ $\sf\green{a = 13}$

We know that,

$\sf\underline{\boxed{\sf a_n = a+(n-1)d } }$

Substituting the found values,

$\sf -17 = 13+(n-1)(-3)$

$\sf -17 = 13-3n+3$

$\sf -17 = 16-3n$

$\sf 3n = 16+17$

$\sf 3n = 33$

$\sf n = 33/3$

$\sf \orange{n = 11}$

$\rule{170}3$