Question

6. The 7th term of an AP is -4 and its 13th term is -16. Find the AP.
... is its first negative term?
ICBSE 2​

Answer 1

Answer:

the AP= 8,6,4,2....

First term (a)=8 (positive)

Step-by-step explanation:

given,

a7= -4

a+(7-1)d = -4

a+6d= -4

therefore, a= -4 - 6d ______eq 1

given,

a13= -16

a + (13-1)d = -16

a+12d= -16

from eq 1 ( a= -4-6d)

-4 -6d +12d = -16

-4 +6d =-16

6d= -16+4

6d = -12

d= -12/6

d= -2

substitute the value of d in eq 1

a= -4 -6x-2

a= -4 +12

a= 8

therefore AP = a,a+1,a+2d....

8,6,4,2,0.....

Answer 2

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

• 7th term = -4
• 13th term = -16

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

• We need to find the AP and first negative term.

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

• 7th term = a + 6d
• 13th term = a + 12d

» a + 6d = -4 .....(i)

» a + 12d = -16 ......(ii)

Solving (i) and ( ii ) we get ,

$\tt a + 6d = - 4 \\ \tt a + 12d = - 16\\ \rm \underline{(-)( - ) \: \: \: \: \: \: ( + ) \: \: } \\ \tt \underline{ \: \: \: \: \: - 6d = + 12} \\ \tt \: \: \: \: \: \: \: d \: \: = - 2$

Putting value of d = -2 in ( i )

↣ a + 6d = -4

↣a + 6 × (-2) = -4

↣a - 12 = -4

↣ a = 8

So, AP :-

» first term (a) = 8

» 2nd term (a+ d) = 8 -2 = 6

» 3rd term (a + 2d) = 8 +(-2)×2 = 4

» 4th term (a + 3d) = 8 + 3 ×(-2) = 2

• ≫AP = 8, 6, 4 ,2.

Now, finding 1st negative term :-

Let the nth term of AP be the 1st negative term.

• Tn < 0

↣[a + (n - 1)d] < 0

↣ [8 + (n - 1)×(-2)]< 0

↣[8 -2n + 2] < 0

↣ [10 - 2n] < 0

↣10 < 2n

↣ n > 5

Therefore n = 6.

Hence,

The 6th term of AP is the first negative term.