Question

5,7,9,.. find the doth term of the arithmetic Sequence ​

Answer 1

ᴛʜᴇ ɢʀᴇᴇᴋ ᴄᴀᴘɪᴛᴀʟ sɪɢᴍᴀ, ᴡʀɪᴛᴛᴇɴ s, ɪs ᴜsᴜᴀʟʟʏ ᴜsᴇᴅ ᴛᴏ ʀᴇᴘʀᴇsᴇɴᴛ ᴛʜᴇ sᴜᴍ ᴏғ ᴀ sᴇǫᴜᴇɴᴄᴇ. ᴛʜɪs ɪs ʙᴇsᴛ ᴇxᴘʟᴀɪɴᴇᴅ ᴜsɪɴɢ ᴀɴ ᴇxᴀᴍᴘʟᴇ:

ᴛʜɪs ᴍᴇᴀɴs ʀᴇᴘʟᴀᴄᴇ ᴛʜᴇ ʀ ɪɴ ᴛʜᴇ ᴇxᴘʀᴇssɪᴏɴ ʙʏ 1 ᴀɴᴅ ᴡʀɪᴛᴇ ᴅᴏᴡɴ ᴡʜᴀᴛ ʏᴏᴜ ɢᴇᴛ. ᴛʜᴇɴ ʀᴇᴘʟᴀᴄᴇ ʀ ʙʏ 2 ᴀɴᴅ ᴡʀɪᴛᴇ ᴅᴏᴡɴ ᴡʜᴀᴛ ʏᴏᴜ ɢᴇᴛ. ᴋᴇᴇᴘ ᴅᴏɪɴɢ ᴛʜɪs ᴜɴᴛɪʟ ʏᴏᴜ ɢᴇᴛ ᴛᴏ 4, sɪɴᴄᴇ ᴛʜɪs ɪs ᴛʜᴇ ɴᴜᴍʙᴇʀ ᴀʙᴏᴠᴇ ᴛʜᴇ s. ɴᴏᴡ ᴀᴅᴅ ᴜᴘ ᴀʟʟ ᴏғ ᴛʜᴇ ᴛᴇʀᴍ ᴛʜᴀᴛ ʏᴏᴜ ʜᴀᴠᴇ ᴡʀɪᴛᴛᴇɴ ᴅᴏᴡɴ.

ᴛʜɪs sᴜᴍ ɪs ᴛʜᴇʀᴇғᴏʀᴇ ᴇǫᴜᴀʟ ᴛᴏ 3×1 + 3×2 + 3×3 + 3×4 = 3 + 6 + 9 + 12 = 30.

3

s 3ʀ + 2

ʀ = 1

ᴛʜɪs ɪs ᴇǫᴜᴀʟ ᴛᴏ:

(3×1 + 2) + (3×2 + 2) + (3×3 + 2) = 24 .

ᴛʜᴇ ɢᴇɴᴇʀᴀʟ ᴄᴀsᴇ

ɴ

s ᴜʀ

ʀ = 1

ᴛʜɪs ɪs ᴛʜᴇ ɢᴇɴᴇʀᴀʟ ᴄᴀsᴇ. ғᴏʀ ᴛʜᴇ sᴇǫᴜᴇɴᴄᴇ ᴜʀ, ᴛʜɪs ᴍᴇᴀɴs ᴛʜᴇ sᴜᴍ ᴏғ ᴛʜᴇ ᴛᴇʀᴍs ᴏʙᴛᴀɪɴᴇᴅ ʙʏ sᴜʙsᴛɪᴛᴜᴛɪɴɢ ɪɴ 1, 2, 3,... ᴜᴘ ᴛᴏ ᴀɴᴅ ɪɴᴄʟᴜᴅɪɴɢ ɴ ɪɴ ᴛᴜʀɴ ғᴏʀ ʀ ɪɴ ᴜʀ. ɪɴ ᴛʜᴇ ᴀʙᴏᴠᴇ ᴇxᴀᴍᴘʟᴇ, ᴜʀ = 3ʀ + 2 ᴀɴᴅ ɴ = 3.

ᴀʀɪᴛʜᴍᴇᴛɪᴄ ᴘʀᴏɢʀᴇssɪᴏɴs

ᴀɴ ᴀʀɪᴛʜᴍᴇᴛɪᴄ ᴘʀᴏɢʀᴇssɪᴏɴ ɪs ᴀ sᴇǫᴜᴇɴᴄᴇ ᴡʜᴇʀᴇ ᴇᴀᴄʜ ᴛᴇʀᴍ ɪs ᴀ ᴄᴇʀᴛᴀɪɴ ɴᴜᴍʙᴇʀ ʟᴀʀɢᴇʀ ᴛʜᴀɴ ᴛʜᴇ ᴘʀᴇᴠɪᴏᴜs ᴛᴇʀᴍ. ᴛʜᴇ ᴛᴇʀᴍs ɪɴ ᴛʜᴇ sᴇǫᴜᴇɴᴄᴇ ᴀʀᴇ sᴀɪᴅ ᴛᴏ ɪɴᴄʀᴇᴀsᴇ ʙʏ ᴀ ᴄᴏᴍᴍᴏɴ ᴅɪғғᴇʀᴇɴᴄᴇ, ᴅ.

ғᴏʀ ᴇxᴀᴍᴘʟᴇ: 3, 5, 7, 9, 11, ɪs ᴀɴ ᴀʀɪᴛʜᴍᴇᴛɪᴄ ᴘʀᴏɢʀᴇssɪᴏɴ ᴡʜᴇʀᴇ ᴅ = 2. ᴛʜᴇ ɴᴛʜ ᴛᴇʀᴍ ᴏғ ᴛʜɪs sᴇǫᴜᴇɴᴄᴇ ɪs 2ɴ + 1 .

ɪɴ ɢᴇɴᴇʀᴀʟ, ᴛʜᴇ ɴᴛʜ ᴛᴇʀᴍ ᴏғ ᴀɴ ᴀʀɪᴛʜᴍᴇᴛɪᴄ ᴘʀᴏɢʀᴇssɪᴏɴ, ᴡɪᴛʜ ғɪʀsᴛ ᴛᴇʀᴍ ᴀ ᴀɴᴅ ᴄᴏᴍᴍᴏɴ ᴅɪғғᴇʀᴇɴᴄᴇ ᴅ, ɪs: ᴀ + (ɴ - 1)ᴅ . sᴏ ғᴏʀ ᴛʜᴇ sᴇǫᴜᴇɴᴄᴇ 3, 5, 7, 9, ... ᴜɴ = 3 + 2(ɴ - 1) = 2ɴ + 1, ᴡʜɪᴄʜ ᴡᴇ ᴀʟʀᴇᴀᴅʏ ᴋɴᴇᴡ.

ᴛʜᴇ sᴜᴍ ᴛᴏ ɴ ᴛᴇʀᴍs ᴏғ ᴀɴ ᴀʀɪᴛʜᴍᴇᴛɪᴄ ᴘʀᴏɢʀᴇssɪᴏɴ

ᴛʜɪs ɪs ɢɪᴠᴇɴ ʙʏ:

sɴ = ½ ɴ [ 2ᴀ + (ɴ - 1)ᴅ ]

ʏᴏᴜ ᴍᴀʏ ɴᴇᴇᴅ ᴛᴏ ʙᴇ ᴀʙʟᴇ ᴛᴏ ᴘʀᴏᴠᴇ ᴛʜɪs ғᴏʀᴍᴜʟᴀ. ɪᴛ ɪs ᴅᴇʀɪᴠᴇᴅ ᴀs ғᴏʟʟᴏᴡs:

ᴛʜᴇ sᴜᴍ ᴛᴏ ɴ ᴛᴇʀᴍs ɪs ɢɪᴠᴇɴ ʙʏ:

sɴ = ᴀ + (ᴀ + ᴅ) + (ᴀ + 2ᴅ) + … + (ᴀ + (ɴ – 1)ᴅ) (1)

ɪғ ᴡᴇ ᴡʀɪᴛᴇ ᴛʜɪs ᴏᴜᴛ ʙᴀᴄᴋᴡᴀʀᴅs, ᴡᴇ ɢᴇᴛ:

sɴ = (ᴀ + (ɴ – 1)ᴅ) + (ᴀ + (ɴ – 2)ᴅ) + … + ᴀ (2)

ɴᴏᴡ ʟᴇᴛ’s ᴀᴅᴅ (1) ᴀɴᴅ (2):

2sɴ = [2ᴀ + (ɴ – 1)ᴅ] + [2ᴀ + (ɴ – 1)ᴅ] + … + [2ᴀ + (ɴ – 1)ᴅ]

sᴏ sɴ = ½ ɴ [2ᴀ + (ɴ – 1)ᴅ]

ᴇxᴀᴍᴘʟᴇ

sᴜᴍ ᴛʜᴇ ғɪʀsᴛ 20 ᴛᴇʀᴍs ᴏғ ᴛʜᴇ sᴇǫᴜᴇɴᴄᴇ: 1, 3, 5, 7, 9, ... (ɪ.ᴇ. ᴛʜᴇ ғɪʀsᴛ 20 ᴏᴅᴅ ɴᴜᴍʙᴇʀs).

s20 = ½ (20) [ 2 × 1 + (20 - 1)×2 ]

= 10[ 2 + 19 × 2]

= 10[ 40 ]

= 400