Question

QUES..ᴛʜᴇ ꜰɪʀꜱᴛ ᴀɴᴅ ᴛʜᴇ ʟᴀꜱᴛ ᴛᴇʀᴍꜱ ᴏꜰ ᴀɴ ᴀᴘ ᴀʀᴇ 10 ᴀɴᴅ 361 ʀᴇꜱᴘᴇᴄᴛɪᴠᴇʟʏ. ɪꜰ ɪᴛꜱ ᴄᴏᴍᴍᴏɴ ᴅɪꜰꜰᴇʀᴇɴᴄᴇ ɪꜱ 9 ᴛʜᴇɴ ꜰɪɴᴅ ᴛʜᴇ ɴᴜᴍʙᴇʀ ᴏꜰ ᴛᴇʀᴍꜱ ᴀɴᴅ ᴛʜᴇɪʀ ᴛᴏᴛᴀʟ ꜱᴜᴍ??? ​

Answer 1

HERE IS YOUR ANSWER ☞

Given, first term, a = 10

Last term, al = 361

And, common difference, d = 9

al = a + (n −1)d

361 = 10 + (n − 1)9

361 = 10 + 9n − 9

361 = 9n + 1

9n = 360

n = 40

Therefore, total number of terms in AP = 40

Now, sum of total number of terms of an AP is given as:

Sn = n/2 [2a + (n − 1)d]

S40 = 40/2 [2 x 10 + (40 − 1)9]

= 20[20 + 39 x 9]

=20[20 + 351]

=20 x 371 = 7420

Thus, sum of all 40 terms of AP = 7420

Yes!! sure we can be friends

can I get your intro?

Answer 2

Question :

ᴛʜᴇ ꜰɪʀꜱᴛ ᴀɴᴅ ᴛʜᴇ ʟᴀꜱᴛ ᴛᴇʀᴍꜱ ᴏꜰ ᴀɴ ᴀᴘ ᴀʀᴇ 10 ᴀɴᴅ 361 ʀᴇꜱᴘᴇᴄᴛɪᴠᴇʟʏ. ɪꜰ ɪᴛꜱ ᴄᴏᴍᴍᴏɴ ᴅɪꜰꜰᴇʀᴇɴᴄᴇ ɪꜱ 9 ᴛʜᴇɴ ꜰɪɴᴅ ᴛʜᴇ ɴᴜᴍʙᴇʀ ᴏꜰ ᴛᴇʀᴍꜱ ᴀɴᴅ ᴛʜᴇɪʀ ᴛᴏᴛᴀʟ ꜱᴜᴍ

Given :

first term (a) = 10

last term ($\tt {T}_{n}$) = 361

Common Difference (d) = 9

To Find :

Number of terms(n) = ?

Total sum ($\tt {S}_{n}$) = ?

Formula :

●$\tt {T}_{n}$ = a + (n-1)d

●$\tt {S}_{n} = \frac{n}{2}$ [2a+(n-1)d]

Solution :

$\tt {T}_{n}$ = a + (n-1)d

$\tt \implies$ 361 = 10 + (n-1)9

$\tt \implies$ 361 = 10 + 9n - 9

$\tt \implies$ 361 = 1 + 9n

$\tt \implies$ 9n = 361 - 1

$\tt \implies$ 9n = 360

$\tt \implies$ n = 360/9

$\tt \implies$ n = 40

Number of terms is 40.

$\tt {S}_{n} = \frac{n}{2}$ [2a+(n-1)d]

$\tt \implies {S}_{n} = \frac{40}{2}$[2(10)+(40-1)9]

$\tt \implies {S}_{n}$ = 20[20+39×9]

$\tt \implies {S}_{n}$ = 20(20 + 351)

$\tt \implies {S}_{n}$ = 20×371

$\tt \implies {S}_{n}$ = 7420

Their sum is 7420.