1 thought on “ques. Tʜᴇ ᴀɴɢʟᴇ ᴏғ ᴇʟᴇᴠᴀᴛɪᴏɴ ᴏғ ᴛʜᴇ ᴛᴏᴘ ᴏғ ᴀ ᴛᴏᴡᴇʀ ғʀᴏᴍ ᴛᴡᴏ ᴘᴏɪɴᴛs ᴅɪsᴛᴀɴᴛ s ᴀɴᴅ ᴛ ғʀᴏᴍ ɪᴛs ғᴏᴏᴛ ᴀʀᴇ ᴄᴏᴍᴘʟᴇᴍᴇɴᴛᴀʀʏ. Pʀᴏᴠᴇ ᴛʜᴀᴛ ᴛʜᴇ”

Answer:

Given, first term, a = 10

361 = 10 + (n − 1)9361 = 10 + 9n − 9

Given, first term, a = 10Last term, al = 361And, common difference, d = 9al = a + (n −1)d361 = 10 + (n − 1)9361 = 10 + 9n − 9361 = 9n + 1

Given, first term, a = 10Last term, al = 361And, common difference, d = 9al = a + (n −1)d361 = 10 + (n − 1)9361 = 10 + 9n − 9361 = 9n + 19n = 360

Given, first term, a = 10Last term, al = 361And, common difference, d = 9al = a + (n −1)d361 = 10 + (n − 1)9361 = 10 + 9n − 9361 = 9n + 19n = 360n = 40

Given, first term, a = 10Last term, al = 361And, common difference, d = 9al = a + (n −1)d361 = 10 + (n − 1)9361 = 10 + 9n − 9361 = 9n + 19n = 360n = 40Therefore, total number of terms in AP = 40

Given, first term, a = 10Last term, al = 361And, common difference, d = 9al = a + (n −1)d361 = 10 + (n − 1)9361 = 10 + 9n − 9361 = 9n + 19n = 360n = 40Therefore, total number of terms in AP = 40Now, sum of total number of terms of an AP is given as:

Given, first term, a = 10Last term, al = 361And, common difference, d = 9al = a + (n −1)d361 = 10 + (n − 1)9361 = 10 + 9n − 9361 = 9n + 19n = 360n = 40Therefore, total number of terms in AP = 40Now, sum of total number of terms of an AP is given as:Sn = n/2 [2a + (n − 1)d]

Given, first term, a = 10Last term, al = 361And, common difference, d = 9al = a + (n −1)d361 = 10 + (n − 1)9361 = 10 + 9n − 9361 = 9n + 19n = 360n = 40Therefore, total number of terms in AP = 40Now, 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]

Given, first term, a = 10Last term, al = 361And, common difference, d = 9al = a + (n −1)d361 = 10 + (n − 1)9361 = 10 + 9n − 9361 = 9n + 19n = 360n = 40Therefore, total number of terms in AP =

Answer:Given, first term, a = 10361 = 10 + (n − 1)9361 = 10 + 9n − 9Given, first term, a = 10Last term, al = 361And, common difference, d = 9al = a + (n −1)d361 = 10 + (n − 1)9361 = 10 + 9n − 9361 = 9n + 1Given, first term, a = 10Last term, al = 361And, common difference, d = 9al = a + (n −1)d361 = 10 + (n − 1)9361 = 10 + 9n − 9361 = 9n + 19n = 360Given, first term, a = 10Last term, al = 361And, common difference, d = 9al = a + (n −1)d361 = 10 + (n − 1)9361 = 10 + 9n − 9361 = 9n + 19n = 360n = 40Given, first term, a = 10Last term, al = 361And, common difference, d = 9al = a + (n −1)d361 = 10 + (n − 1)9361 = 10 + 9n − 9361 = 9n + 19n = 360n = 40Therefore, total number of terms in AP = 40Given, first term, a = 10Last term, al = 361And, common difference, d = 9al = a + (n −1)d361 = 10 + (n − 1)9361 = 10 + 9n − 9361 = 9n + 19n = 360n = 40Therefore, total number of terms in AP = 40Now, sum of total number of terms of an AP is given as:Given, first term, a = 10Last term, al = 361And, common difference, d = 9al = a + (n −1)d361 = 10 + (n − 1)9361 = 10 + 9n − 9361 = 9n + 19n = 360n = 40Therefore, total number of terms in AP = 40Now, sum of total number of terms of an AP is given as:Sn = n/2 [2a + (n − 1)d]Given, first term, a = 10Last term, al = 361And, common difference, d = 9al = a + (n −1)d361 = 10 + (n − 1)9361 = 10 + 9n − 9361 = 9n + 19n = 360n = 40Therefore, total number of terms in AP = 40Now, 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]Given, first term, a = 10Last term, al = 361And, common difference, d = 9al = a + (n −1)d361 = 10 + (n − 1)9361 = 10 + 9n − 9361 = 9n + 19n = 360n = 40Therefore, total number of terms in AP =