{"id":7049,"date":"2021-10-22T15:55:39","date_gmt":"2021-10-22T10:25:39","guid":{"rendered":"https:\/\/mathemerize.com\/?p=7049"},"modified":"2021-10-25T02:19:41","modified_gmt":"2021-10-24T20:49:41","slug":"a-man-saves-rs-200-in-each-of-the-first-three-months-of-his-service-in-each-of-the-subsequent-months-his-saving-increases-by-rs-40-more-than-the-saving-of-immediately-previous-month-his-total-savin","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/a-man-saves-rs-200-in-each-of-the-first-three-months-of-his-service-in-each-of-the-subsequent-months-his-saving-increases-by-rs-40-more-than-the-saving-of-immediately-previous-month-his-total-savin\/","title":{"rendered":"A man saves Rs 200 in each of the first three months of his service. In each of the subsequent months, his saving increases by Rs 40 more than the saving of immediately previous month. His total saving from the start of service will be Rs 11040 after"},"content":{"rendered":"

Solution :<\/h2>\n

Let the time taken to save Rs 11040 be (n + 3) months.<\/p>\n

for first three months, he saves Rs 200 each month.<\/p>\n

In (n + 3) months,<\/p>\n

3 \\(\\times\\) 200 + \\(n\\over 2\\) { 2(240) + (n – 1) \\(\\times\\) 40 } = 11040<\/p>\n

\\(\\implies\\)\u00a0 600 + \\(n\\over 2\\) {40(12+ n – 1)} = 11040<\/p>\n

\\(\\implies\\)\u00a0 600 + 20n(n + 11) = 11040<\/p>\n

\\(\\implies\\) 30 + \\(n^2\\) + 11n = 552<\/p>\n

\\(\\implies\\)\u00a0 \\(n^2\\) + 11n – 552 = 0<\/p>\n

\\(\\implies\\)\u00a0 \\(n^2\\) + 29n – 18n – 552 = 0<\/p>\n

\\(\\implies\\)\u00a0 n(n + 29)\u00a0 – 18(n +29) = 0<\/p>\n

\\(\\implies\\)\u00a0 \u00a0(n – 18)(n + 29) = 0<\/p>\n

\\(\\therefore\\)\u00a0 \u00a0n = 18,\u00a0 n = -29 (neglect)<\/p>\n

\\(\\therefore\\)\u00a0 \u00a0Total time = (n + 3) = 21 months<\/p>\n

\n

Similar Questions<\/h3>\n

Let \\(a_n\\) be the nth term of an AP. If \\(\\sum_{r=1}^{100}\\) \\(a_{2r}\\) = \\(\\alpha\\) and \\(\\sum_{r=1}^{100}\\) \\(a_{2r-1}\\) = \\(\\beta\\), then the common difference of the AP is<\/a><\/p>\n

If \\((10)^9\\) + \\(2(11)^1(10)^8\\) + \\(3(11)^2(10)^7\\) + \u2026\u2026 + \\(10(11)^9\\) = \\(K(10)^9\\), then k is equal to<\/a><\/p>\n

If 100 times the 100th term of an AP with non-zero common difference equal to the 50 times its 50th term, then the 150th term of AP is<\/a><\/p>\n

If x, y and z are in AP and \\(tan^{-1}x\\), \\(tan^{-1}y\\) and \\(tan^{-1}z\\) are also in AP, then<\/a><\/p>\n

The sum of first 20 terms of the sequence 0.7, 0.77, 0.777, \u2026\u2026. , is<\/a><\/p>\n","protected":false},"excerpt":{"rendered":"

Solution : Let the time taken to save Rs 11040 be (n + 3) months. for first three months, he saves Rs 200 each month. In (n + 3) months, 3 \\(\\times\\) 200 + \\(n\\over 2\\) { 2(240) + (n – 1) \\(\\times\\) 40 } = 11040 \\(\\implies\\)\u00a0 600 + \\(n\\over 2\\) {40(12+ n – …<\/p>\n

A man saves Rs 200 in each of the first three months of his service. In each of the subsequent months, his saving increases by Rs 40 more than the saving of immediately previous month. His total saving from the start of service will be Rs 11040 after<\/span> Read More »<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"default","ast-global-header-display":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","theme-transparent-header-meta":"default","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":""},"categories":[43,57],"tags":[],"yoast_head":"\nA man saves Rs 200 in each of the first three months of his service. In each of the subsequent months, his saving increases by Rs 40 more than the saving of immediately previous month. 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His total saving from the start of service will be Rs 11040 after\" \/>\n<meta property=\"og:description\" content=\"Solution : Let the time taken to save Rs 11040 be (n + 3) months. for first three months, he saves Rs 200 each month. In (n + 3) months, 3 (times) 200 + (nover 2) { 2(240) + (n – 1) (times) 40 } = 11040 (implies)\u00a0 600 + (nover 2) {40(12+ n – … A man saves Rs 200 in each of the first three months of his service. In each of the subsequent months, his saving increases by Rs 40 more than the saving of immediately previous month. 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