Question

7. The first term of an A.P. is 5, the last term is
45 and the sum of its terrns is 1000. Find the
number of terms and the common difference
of the À.P.​

Answer 1

Answer:

n=40              d=40/39

Step-by-step explanation:

a be first term           l be last term

SUM= n/2(a+l)

1000=25n

n=40

l=a+(40-1)d

45=5+39d

d=40/39

Answer 2

[HeY Mate]

$\huge\bold\red{Answer}$

Given,

• First term of AP (a) = 5.
• Last term (l) = 45.
• Sum of terms (Sn) = 1000.

.

.

To Find,

• No. of terms (n) = ?
• Common difference (d) = ?

.

.

Solution:

First, we will find 'n' (no. of terms).

By using formula of Sn,

$\tt S_{n} \implies \frac{n}{2} (a + l) \\ \tt 1000 = \frac{n}{2} \times (5 + 45) \\\tt 1000 = 25n \\ \tt n = \frac{1000}{25} \\ \tt n = 40 \\ \\ \\ </p><p>$

Now,

d is still unfamiliar,

By using another formula of Sn,

$</strong><strong>\tt</strong><strong> </strong><strong>s_{n} </strong><strong>\</strong><strong>i</strong><strong>m</strong><strong>p</strong><strong>l</strong><strong>i</strong><strong>e</strong><strong>s</strong><strong> \frac{n}{2} (2a + n - 1)d \\ </strong><strong>\tt</strong><strong> </strong><strong>1000= \frac{40}{2} (2 \times 5 + 40 - 1)d \\ </strong><strong>\tt</strong><strong> </strong><strong>1000 = 20(</strong><strong>1</strong><strong>0</strong><strong>+</strong><strong> </strong><strong>3</strong><strong>9d) \\</strong><strong>\tt</strong><strong> </strong><strong>3</strong><strong>9</strong><strong>d</strong><strong> = </strong><strong>5</strong><strong>0</strong><strong>-</strong><strong>1</strong><strong>0</strong><strong> \\ </strong><strong>\tt</strong><strong> </strong><strong>d</strong><strong> = \frac{</strong><strong>4</strong><strong>0</strong><strong>}{</strong><strong>3</strong><strong>9</strong><strong>} </strong><strong>$

I Hope It Helps You✌️